


PLANE TRIGONOMETRY 

AND 

^PLICATIONS 




Class , 

Book. 

Copyright^?, 



COPYRIGHT DEPOSHV 



PLANE TRIGONOMETRY 

AND 

APPLICATIONS 



BY 

E. J. WILCZYNSKI, Ph.D. 
it 

THE UNIVERSITY OP CHICAGO 



EDITED BY 

H. E. SLAUGHT, Ph.D. 

THE UNIVERSITY OF CHICAGO 



ALLYN and BACON 
Boston Neto gorfc Chicago 






COPYRIGHT, 1914, BY 
E. J. WILCZYNSKI 



NorfoooU Press 

J. S. Cushing Co. — Berwick & Smith Co. 

Norwood, Mass., U.S.A. 

■FEB -71914 

©CI.A36 2491 



PREFACE 

The characteristic features of this book may be summarized as 
follows : 

1. The method of presentation is thoroughly heuristic. This en- 
ables the student to get a firm grasp of the subject by teaching 
him to recognize the fundamental ideas which underlie and unify 
the separate steps of the mathematical argument, instead of con- 
fusing him by a disconnected dogmatic statement of isolated facts. 

2. The book is divided into two parts. The first part is devoted 
to the theoretical and numerical solution of triangles. The 
second part treats of the functions of the general angle, their 
addition theorems and other properties, together with applica- 
tions to simple harmonic curves, simple harmonic and wave 
motion, and harmonic analysis. Part One is also published sepa- 
rately and is well adapted for use in secondary schools. The 
complete book is intended for the freshman course in colleges. 

3. TJie discussion of the solution of triangles, in Part One, is not 
interrupted by any digressions about coordinate systems, addition 
theorems, and the like. It has been thought desirable to post- 
pone to the second part the consideration of all of these matters, 
which are indeed important but unnecessary for the solution of 
triangles. 

4. The definitions for the functions of an obtuse angle have been 
made to grow organically out of the needs of the problem of solv- 
ing triangles, in a way which seems both simple and natural, and 
which at the same time illustrates an important principle of 
mathematical procedure. 

5. Tlie ichole theory of triangles has been unified by giving a 
central position to the area problem. As a consequence, almost 
all of the necessary equations present themselves spontaneously 
and in a connected fashion. The law of tangents is the only one 
which causes any difficulty in this respect. But the law of tan- 
gents also has been made to submit to a heuristic treatment, by 



iv PREFACE 

introducing the notion of the form-ratio of a triangle, and combin- 
ing this notion with a direct geometric proof of the formulae for 
sin A + sin B and sin A — sin B. 

6. Tlie numerical aspect of the work has been discussed very fully. 
Directions for computation are given in great detail ; most of the 
common sources of error are pointed out ; and methods for detect- 
ing and correcting them are indicated. After a thorough discus- 
sion of the significance of the number of decimal places needed in 
a computation, the student is urged to train and use his judgment 
on this matter. He is given an opportunity to do this by supply- 
ing him with complete five- and three-place tables and a partial 
set of four-place tables. 

7. The slide rule is explained with considerable detail and its 
use recommended. A number of other labor-saving devices are 
discussed. 

8. The examples have been selected with great care. Examples 
without real significance have been avoided, and the numbers 
have been chosen so as not to lead to five-place calculations when 
such a show of accuracy would be absurd. Special efforts have 
been made to word the examples in such a way as to avoid 
ambiguity. 

9. The applications cover a wider field than usual, and include 
problems in heights and distances, surveying, navigation, engi- 
neering, astronomy, and physics. But the examples involving 
such applications are not, as in most texts, introduced at random 
and without previous explanation. Every notion which is re- 
quired for the solution of any example in the book is fully 
explained on the spot or in some earlier portion of the text. 

10. The use of a few new terms, such as the standard position of 
an angle, odd and even cardinal angles, has helped to simplify 
materially the statement of a number of important results. 

11. The addition formulas are presented in two different ways. 
The first, more elementary method, is made to yield the general 
result by the help of mathematical induction. The second 
method, based on the notions of directed lines, line-segments, and 
angles, appears here in a very simple and elegant form. 

12. The articles on harmonic and wave motion tend to show the 
student fchat Trigonometry has other applications besides the 
solution of triangles. 



PREFACE v 

13. Aconsiderable amount of historical matter has been introduced, 
not in the form of detached historical notes, but organically con- 
nected with the topic under discussion. Most of this matter was 
gathered from Braunmuhl's Vorlesungen uber die Geschichte der 
Trigonometric Professors Cajori and Karpinski have kindly 
answered some questions of a historical nature about which we 
were in doubt. 

14. The type and the manner of spacing used in the tables are 
the results of a number of experiments, the object being to pro- 
duce a set of tables which should be as pleasant to the eye as 
possible. The tables are bound separately for various reasons. 
In order to make them easily legible, a certain size of page was . 
necessary, and it was thought undesirable to use so large a page 
for the text itself. In the second place, it is a great advantage 
for the student if he can have his text and his tables open before 
him at the same time. In the third place, it is often desirable, 
in examinations, to allow students to use their tables without 
their books. Finally, a separation of the tables and text makes 
it easy to use this text with other tables, or these tables with 
other texts, thus providing a maximum of elasticity in organizing 
a course. 

Many of the .older texts on Trigonometry have been consulted 
during the preparation of this book, and the attempt has been 
made to learn from all of them. The works of Serret, Lubsen, 
Wiegaxd, Crockett, Moritz, Hall and Frink have been espe- 
cially helpful. A few purely numerical examples have actually 
been taken from these and other texts without change, so as to 
reduce somewhat the task of computing the answers, and at the 
same time to make the answers more trustworthy. Most of the 
examples, however, are new ; many of them are new in kind. 

In conclusion, the author and editor wish to acknowledge their 
indebtedness to their colleagues at the University of Chicago for 
various helpful suggestions and criticisms. 

E. J. WILCZYNSKI. 

H. E. SLAUGHT, Editor. 



CONTENTS 

PART ONE. SOLUTION OF TRIANGLES 

CHAPTER I 

The Object of Trigonometry. 

PAGE 

1. Direct measurement of lines ....... 1 

2. -Direct measurement of angles 3 

3. The impossibility of finding all distances by direct measure- 

ment 6 

4. The graphic method 8 

5. The desirability of an arithmetical method for solving triangles 9 



CHAPTER II 

The Trigonometric Functions of Acute Angles. 

6. The necessity of introducing new ideas 11 

7 . Definition of the trigonometric functions of an acute angle . 13 

8. The practical need of tables giving the values of the trigo- 

nometric functions ........ 16 

9. Relations between the six trigonometric functions of an acute 

angle 18 

10. Relations between the functions of complementary angles . 20 

11. The values of the functions of 0°, 30°, 45°, 60°, 90° . . . 22 



CHAPTER III 

Solution of Right Triangles by Natural Functions. 

12. Arrangement and use of the table of natural functions . . 25 

13. Solution of right triangles by means of the table of natural 

functions ... 20 



viii CONTENTS 

CHAPTER IV 

PAGB 

Discussion of Some Devices for reducing the Labor involved in Numer- 
ical Computations. 

14. The number of decimal places 34 

15. The accuracy of a sum, difference, product, or quotient of two 

numbers obtained by measurement 36 

16. Labor-saving devices 37 

17. Definition of logarithms 38 

18. The properties of logarithms 43 



CHAPTER V 



Calculation by Logarithms. 



19. Common logarithms 47 

20. Properties of the mantissa 48 

21. Determination of the characteristic 49 

22. Arrangement and use of the table of logarithms . . . . 51 

23. Cologarithms 54 

24. Extraction of roots by means of logarithms .... 55 

25. Logarithmic calculations which involve negative numbers . 56 

26. The logarithms of the trigonometric functions . . . .57 

27. The accuracy of five-place tables 59 

28. The trigonometric functions of angles near 0° or 90° . . 60 

29. The logarithmic or Gunter scale 61 

30. The slide rule 62 



CHAPTER VI 

Application of Logarithms to the Solution of Right Triangles. 

31. The general method 67 

32. The preliminary graphic solution 67 

33. The gross errors 68 

34. Selection of formulse and checks 68 

35. The framework or skeleton form 69 

36. The computation ' . .70 

37. Revision of the computation when the check is unsatisfactory 70 

38. Applications to simple problems of surveying, navigation, and 

geography .......... 72 

39. Right triangles of unfavorable dimensions .... 80 



CONTENTS ix 

CHAPTER VII 

PAGE 

Theory of Oblique Triangles. 

40. The area of an oblique triangle in terms of two of its sides and 

the included angle .... 

41. The law of sines 

42. The law of cosines. A generalization of the theorem of 

Pythagoras 

43. Properties of the functions of an obtuse angle 

44. Other formulae for the area of an oblique triangle 
4-3. The radius and center of the inscribed circle 

46. The half-angle formulae . . 

47. The circumscribed circle .... 

48. The form ratios of a triangle . 

49. The formula? for the sum and difference of two 

50. The law of tangents ..... 

51. A second proof of the law of tangents and Mollw 



sines 

100 
eide's equations 102 



82 
84 

85 
88 
90 
93 
95 
95 
97 



CHAPTER VIII 

Solution of Oblique Triangles. 

52. The fundamental problem ....... 107 

53. Case I. Given one side and two angles 108 

54. Case II. Given two sides and the included angle . , . 109 

55. Case III. Given two sides and the angle opposite to one of them 111 

56. Case IV. Given the three sides of the triangle . . .116 

57. Problems in heights and distances, plane surveying, and plane 

sailing 118 

58. Displacements, velocities, and forces 126 

59. Reflection and refraction of liuht 128 



PART TWO. PROPERTIES OF THE TRIGONOMETRIC 
FUNCTIONS 

CHAPTER IX 

The General Angle and its Trigonometric Functions. 

60. The notion of the general angle 133 

61. Initial and terminal side. Standard position of an angle . 135 

62. The notion of the trigonometric functions of a general angle . 136 

63. Rectangular coordinates 137 



CONTENTS 

PAGE 

64. Definition of the trigonometric functions of a general angle . 140 

65. Discussion of the exceptional cases 141 

66. The four quadrants 145 

67. General character of the trigonometric functions. Their 

periodicity 146 

68. Relations between the trigonometric functions of a general 

angle 148 

69. Trigonometric identities which involve functions of a single 

angle 149 



CHAPTER X 

Graphic Representations of the Trigonometric Functions. 

70. Line representation of the trigonometric functions . . . 151 

71. Graphs of functions, a number of whose numerical values are 

given 155 

72. Graphs of simple algebraic functions 158 

73. Graphs of the trigonometric functions ..... 159 

74. The natural unit of circular measurement. Definition of a 

radian 163 

75. Relations between the functions of two symmetrical angles . 167 

76. Relations between functions of two angles whose sum or differ- 

ence is a right angle 169 

77. The quadrantal formulss 171 

78. Properties of the sine and cosine curves 175 



CHAPTER XI 

Relations between the Functions of More than One Angle. 

79. The addition theorems for sine and cosine .... 180 

80. The addition theorems for tangent and cotangent . . . 185 

81. The subtraction formulae 186 

82. Formula for converting products of trigonometric functions 

into sums, and vice versa 187 

83. Functions of double angles 189 

84. Functions of half angles . . . . . . . . 190 

85 a. The limit ^-^ and related limits 193 

6 

85 6. The auxiliary quantities 8 and T 198 



CONTENTS xi 
CHAPTER XII 

PAGE 

Directed Lines and Directed Line-segments. 

86. Plan of another proof for the addition formulae . . .201 

87. Directed lines and segments 201 

88. Angles between directed lines 204 

89. Projections 204 

90. Projection of a broken line 207 

91. Direction cosines of a line 209 

92. Formula for the cosine of the angle between two lines whose 

direction cosines are given 210 

93. New proof for the addition and subtraction formulae . . 212 

94. The Generalized law of sines 212 



CHAPTER XIII 

The Inverse Trigonometric Functions and Trigonometric Equations. 

95. The problem of inverting the trigonometric functions . .214 

96. Determination of all of the angles which correspond to a given 

value of one of the functions 214 

97. The inverse trigonometric functions . . . . . 217 

98. Trigonometric equations 222 

99. The equation a sin 6 + b cos 6 = c 223 



CHAPTER XIV 

Applications to the Theory of Wave Motion. 

100. Simple harmonic motion 226 

101. The period and phase constant 228 

102. Some illustrations of simple harmonic motion . . . 229 

103. Simple harmonic curves 231 

104. Amplitude 232 

105. Wave length 232 

106. Phase constant 233 

107. Wave motion 235 

108. General harmonic motion 239 

109. General harmonic curves ....... 241 

110. Harmonic analysis or trigonometric interpolation . . . 243 

111. Theorems leading to the general solution of the problem of 

trigonometric interpolation 249 

112. General solution of the problem of trigonometric interpolation 257 



PLANE TRIGONOMETRY 

AND 

APPLICATIONS 

PART ONE 

SOLUTION OF TRIANGLES 
CHAPTER I 

THE OBJECT OF TRIGONOMETRY 

1. Direct measurement of lines. One of the most com- 
mon operations of practical geometry is that of measuring 
the distance between two points. In its simplest form this 
consists merely in the repeated application of some unit of 
measurement to the required distance. 

The units of measurement most frequently used for this purpose 
are a foot rule, a yardstick, a surveyor's chain, tape lines of definite 
length, etc. Fractional parts of the unit are usually read from a 
graduated scale, engraved or stamped on the standard used. A familiar 
illustration of this is the scale of inches on an ordinary foot rule. 

In spite of the apparent simplicity of this process, it is a 
matter of great practical difficulty to carry out such meas- 
urements with a high degree of precision. The sources of 
error are numerous and, in part, unavoidable. No instru- 
ment made by man is absolutely accurate. Thus, if we use 
a yardstick, it will not be absolutely straight, and it may be 
a trifle too long or too short. It will be very difficult to 
make sure that we are laying off the second yard of our dis- 
tance exactly where the first yard ends. Consecutive posi- 
tions of the yardstick will form angles with each other, 
which are not exactly equal to 180°. In fact, it is almost 
impossible to run a straight line of considerable length, with 

l 



2 THE OBJECT OF TRIGONOMETRY 

any degree of accuracy, without the help of more complicated 
instruments, such as the transit described in Art. 2. 

The graduated scales, used for measuring fractional parts 
of the unit of length, are also affected by various sources of 
inaccuracy, and it will be difficult for the observer to esti- 
mate accurately a fractional part of the smallest visible divi- 
sion on the scale. 

Enough has been said to indicate just a few of the many 
difficulties encountered in the, apparently so simple, opera- 
tion of measuring the length of a line, and to emphasize the 
fact that we must always regard the result of such a meas- 
urement as an approximation, even if the most refined instru- 
ments known to Science have been used. 

The difference between rough and fine measurements is one of degree 
only. The more refined the method, the smaller will be the " probable 
error " and the closer the approach to the truth. But we can never be 
sure that a quantity has been measured with absolute precision. 

EXERCISE I 

1. What are some of the sources of inaccuracy in measuring the 
length of a table? 

2. If you wish to measure the distance diagonally across a table, by 
means of a foot rule, what additional sources of inaccuracy will appear ? 
Would a stretched cord be of some use in this connection ? 

3. How would you measure the distance diagonally across a room 
from one of the floor corners to the opposite corner of the ceiling? Do 
you know of any other method by which this distance might be obtained, 
except that of direct measurement ? 

4. How would you join two given points by a straight line (say for 
the purpose of constructing a fence), over a level piece of ground, if the 
distance is too great to enable you to stretch a cord? Do you know of 
any property of the line of sight which might help in the solution of 
such a problem ? 

5. If you attempt to measure two different distances, of which one 
is about ten times as great as the other, using the same foot rule and the 
same method of measurement in both cases, which of these two dis- 
tances will probably be obtained with greater accuracy ? Why ? 

6. What difficulties arise, and how would you attempt to meet them, 
if you were asked to measure the horizontal distance between two points 
on an uneven piece of ground ? 



DIRECT MEASUREMENT OF ANGLES 3 

7. Suppose you have measured a distance (say the edge of a table) 
with great care and have found it to be equal to -i feet and 9f inches. 
Is this the exact length of the table, or may it be a small fraction of an 
inch greater or less ? 

8. If you were to repeat the measurement with still greater care, 
making use of a more perfect standard of length, is it likely that you 
would find exactly the same result as before ? 

9. In any such measurement can you ever attain absolute accuracy? 
If not, why not ? 

10. Is there any way of knowing whether a measurement is absolutely 
accurate ? Is there any way of knowing whether it is accurate to within 
a certain limit of accuracy, say one tenth of an inch ? 

11. If a distance has been measured by a process which may be relied 
upon to give a result accurate only to one one-hundredth of an inch, is 
it desirable, proper or honest, to give the result expressed in decimal 
parts of an inch to more than two decimal places? Why not? In per- 
forming calculations based upon such measurements, how many decimal 
places should we ordinarily carry? 

12. In measuring distances by means of metal rods, when great ac- 
curacv is required, changes of temperature must be taken into account. 
Why? 

13. With what units of length are you familiar ? 

14. Gather from an encyclopedia what you can concerning the 
" standard yard " kept at Washington. 

15. What is the metric system? What are the relations to each 
other of the units called millimeter, centimeter, decimeter, and meter? 
What is the length of a meter in inches ? 

16. Can you see any reason why the metric system should be prefer- 
able to the English system of weights and measures? 

2. Direct measurement of angles. The operation of measur- 
ing angles is scarcely less important than that of measuring 
distances. A pro- 
tractor is an instru- 
ment used for this 
purpose. In its 
simplest form, a pro- 
tractor consists of 
an arc of a circle 
graduated to de- 
grees (Fig. 1). A 




Fig. 1 



THE OBJECT OF TRIGONOMETRY 



mere inspection of the instrument will enable the student 
to see how angles may be measured and constructed by means 
of it. 

The unit usually employed in measuring angles is the 
ninetieth part of a right angle, and is called a degree. A 
degree is divided into sixty equal parts, each of which is called 
a minute, and each minute is subdivided into sixty seconds. 
Thus 

60 seconds (60") = one minute (l')» 

60 minutes (60') = one degree (1°), 
90 degrees (90°) = one right angle. 

Very frequently, the angles smaller than one degree are 
expressed as decimal parts of a degree instead of in minutes 

and seconds. 

For the purpose of 
measuring angles in 
the field, surveyors 
make use of an in- 
strument called a 
transit or theodolite. 
The essential parts 
of this instrument 
are (cf . Fig. 2) : 

1. a horizontal 
graduated circle ; 

2. a movable cir- 
cular plate adjusted 
so as to be capable of 
rotation around the 
center of the hori- 
zontal graduated cir- 
cle ; 

3. an index at- 
tached to the movable 
plate in such a way 

p IG . 2 as to enable the ob- 




DIRECT MEASUREMENT OF ANGLES 5 

server to read off the amount of its rotation with reference to 
the fixed horizontal circle ; 

4. two standards attached to the movable plate and carry- 
ing a horizontal axis to which is attached a telescope and 
also, in a complete instrument, a vertical graduated circle, 
used for measuring angles whose sides lie in a vertical plane. 

The transit is usually supported on a tripod. If we wish to measure 
the angle between two horizontal lines, the tripod is placed over the 
vertex of the angle and the telescope is pointed toward some point on 
one of the sides of the angle. The index will then point to a definite 
division on the horizontal circle. The operation of ascertaining the 
division of the circle toward which the index is pointing, is known as 
"reading the circle." After reading the circle and noting the result, the 
telescope is directed toward a point on the other side of the angle. The 
difference between the two readings of the horizontal circle will give 
the magnitude of the angle. 

In a similar way the vertical circle, which is attached to the axis of 
the telescope, makes possible the measurement of angles whose sides lie 
in a vertical plane. 

Both circles are usually graduated to whole degrees. The index, in 
most instruments, is not a simple pointer, but a so-called vernier, an in- 
genious device which enables the .observer to determine the reading of 
the circle to within a small fraction of a degree, even if the circle is 
graduated only to whole degrees. In the most accurate instruments, the 
vernier is replaced by a reading microscope. 

It is obviously very important that the horizontal circle be exactly 
horizontal, and that its center be exactly over the vertex of the angle 
which is to be measured. To help in making these adjustments, the 
surveyor uses a spirit level and a plumb line. 

Most transits are also supplied with a compass, which enables the 
observer to determine the absolute directions of the lines which he is 
surveying. 

EXERCISE II 

1. Use a protractor to draw angles of 10°, 20°, 31°, 47° 30', 67°, 
78°, 86°. 

2. Draw five angles at random and measure them as accurately as 
possible with your protractor. 

3. Draw a triangle at random, measure its angles and find their 
sum. What should this sum be ? If you have obtained a different result 
for the sum, what are the reasons ? 



6 THE OBJECT OF TRIGONOMETRY 

4. Construct, out of cardboard, an instrument embodying the prin- 
ciple of the transit, substituting for the telescope some other method of 
taking a sight. 

5. Why is it important that the horizontal circle of a transit should 
be truly horizontal? How does the spirit level enable us to make it so? 
Study the article on the spirit level in some encyclopedia or in a treatise 
on surveying. 

6. Study the articles on vernier, micrometer, reading microscope, 
compass, in an encyclopedia or in some appropriate treatise, and write an 
abstract of the same. 

7. Describe the sources of inaccuracy which you think may arise in 
the measurement of an angle by means of a protractor or theodolite. 

8. Deduce rules for converting minutes and seconds into decimal 
parts of a degree and vice versa. 

9. Apply this rule to the angles 

30° 20', 10° 45', 8° 40' 20", 3° 8' 2". 

10. Express the following angles in degrees, minutes, and seconds : 

23°.14, 18°.25, 46°.235. 

3. The impossibility of finding all distances by direct 
measurement. We have discussed briefly the direct methods 
for measuring distances and angles. It is clear from our 
discussion that it is very difficult to measure great distances 
in that way. But there are many cases in which it is al- 
together impossible to apply such direct methods. How, for 
example, should we proceed to find the distance through a 
mountain or across an extensive valley? Ho\g shall we find 
the distance from New York to London, or from the earth to 
the moon, by direct measurement ? 

It is clear that, if we wish to answer such questions at all, 
we shall have to devise some method different from that of 
direct measurement. 

The attempt to solve, by indirect methods, problems whose direct 
solution is inconvenient or impossible usually leads to great advances in 
Science. The most important theorems of elementary geometry were 
probably first discovered by the Ancients in their attempts to devise con- 
venient and practical methods for the measurements which they found 
necessary for the purposes of their everyday life. It is generally believed, 
for instance, that the Egyptians became expert geometricians and sur- 
veyors at an early period of their history, because it was so important 



DIFFICULTY OF DIRECT MEASUREMENT 7 

for them to be able to reestablish the boundary lines of their lands after 
the efface men t of their marks by the annual inundations of the Nile. 

The earliest Greek philosophers and mathematicians were pupils of 
the Egyptians, and some of their first achievements were connected with 
problems of the particular kind which we are now discussing. Thus it 
is reported that Tiialks of Miletus (about 600 b.c.) measured the 
height of a pyramid by measuring the length of its shadow, at the instant 
when the shadow of a vertical stick by its side was exactly asjong as the 
stick itself. The height of the pyramid would then be equal 'to the 
length of its shadow at that moment. 

This method has the inconvenient feature of compelling the observer 
to wait (many hours perhaps) for the right moment. According to a 
report by Plutarch, Thales also devised a second method which 
avoided this inconvenience. This method involves the use of the simplest 
properties of similar triangles and may be illustrated by means of the 
following example : 

We place a stick 5 feet high into the ground near a building whose 
height we wish to find. At any convenient moment we measure the 




125 feet 




Fig. 



length of the shadows of the stick and of the building. Suppose we 
find in this way that the shadow of the building is 125 feet long at the 
moment when the shadow of the stick is 10 feet long. If h denotes the 
height of the building, we shall have (Fig. 3) 



whence 



h : 125 = 5 : 10, 
h = 62.5 feet. 



According to Eude.mus (about 300 b.c), one of the earliest writers 
on the history of mathematics, Thales also invented a method for 
measuring the distance from the shore to 
a ship at sea. Although Eudemus does 
not describe Thales's method in detail, he 
says enough to lead us to conclude that it 
was essentially as follows : Let BL be a 
tower, say a lighthouse on the shore, and Fig. 4 




8 THE OBJECT OF TRIGONOMETRY 

let £ be the ship at sea. Measure the angle BBS and the height of the 
tower. Construct a similar triangle B'L' S' on the drawing board and 
measure B'S' and B'L'. Then BS can be found from the proportion 
BS:BL = B'S , :B'L'. 

4. The graphic method. The examples discussed in Art. 
3 show how we may solve many problems of practical geom- 
etry by indirect measurements. We did not measure directly 
the quantity which we were seeking, but some other related 
quantities, and ultimately, by means of these relations, we 
determined the desired quantity itself. 

But these same examples may serve to illustrate another 
point. The solutions which we gave are examples of the 
graphic method, that is, of a process which makes use of 
drawing instruments, of accurate geometric constructions 
and measurements, for the purpose of obtaining the values 
of the unknown quantities. Such graphical methods are 
often extremely valuable and have been developed in recent 
times, in connection with other parts of mathematics, so as 
to give rise to very important results. 

The graphic solution of any problem about triangles will finally re- 
duce to an application of certain theorems of geometry which state that 
a triangle is determined and may be constructed when certain ones of its 
six parts (sides and angles) are given. It is clear, then, that these 
theorems are particularly important for the graphic method. Some of 
the following questions have been chosen for the purpose of aiding the 
student to refresh his memory in regard to these matters. 

EXERCISE III 

In the following examples and throughout the book we shall usually 
denote the angles of a triangle by A, B, C and the sides opposite to these 
angles by a, b, c, respectively. The student should be provided with a 
protractor, a pair of compasses, and a ruler divided decimally, say into 
centimeters and millimeters. All constructions and measurements should 
be made as carefully as possible. 

1. Given a = 3.72, b = 4.91, c = 2.56. Find the angles. 

2. Given a = 4.27, B = 35°, C = 69°. Find the remaining parts of 
the triangle. 

3. Given b = 5.63, c = 6.71, A = 27°. Find the remaining parts of 
the triangle. 



AN ARITHMETICAL METHOD 9 

4. Given a = 4.2:3, b = 5.16, A = 55°. Find the remaining parts of 
the triangle. 

5. When a, b, c are given at random, can we always find a triangle 
of which a. &, c are the sides, or is there some restriction on the possible 
values of a, b, c? 

6. If a, b, A are given, there may be one or two solutions, or no 
solution. Discuss these cases. 

7. Is a triangle determined when we know the magnitudes of its three 
angles? Why? When the three angles of a triangle are given, have we 
obtained essentially more information than if only two of them are 
given? Why? Is the third angle of a triangle independent of the other 
two? 

8. What do you mean by similar triangles? 

9. Under what conditions are two triangles similar? 

10. Use the shadow method of Thales (Art. 3) to measure the height 
of some building in your neighborhood. 

11. By means of a transit, or else by means of the home-made in- 
strument of cardboard suggested in Ex. 4 of Exercise II, find the dis- 
tance of some object situated on the opposite side of the street from your 
home, without crossing the street. Afterward check your result by 
direct measurement. 

12. How may a person on board ship find his distance from a build- 
ing on the shore if he knows its height? 

5. The desirability of an arithmetical method for solving 

triangles. We have seen that it is a rather simple matter to 
solve a triangle by the graphic method. But we can hardly 
feel altogether satisfied with the graphic solution, for it will 
clearly not permit us to reach any great degree of accuracy. 
To be sure, by making our drawings on a very large scale, 
we might lay off distances with considerable precision, but 
we should still encounter the difficult}^ of accurately plotting 
angles. Clearly it would require extraordinary skill and 
exceedingly fine instruments to enable us to draw an angle 
so accurately that its error should not exceed one minute of 
arc. Many other circumstances combine to make a graphical 
solution unsatisfactory if a high degree of precision is required. 

The main value of the graphic method lies in furnishing a 
solution whose approximate correctness is apt to he apparent to 



10 THE OBJECT OF TRIGONOMETRY 

the eye, and which may therefore, in almost all cases, serve as a 
check on the more complete solution obtained in some other way. 

But the lack of accuracy is only one of the defects of the 
graphic method, although perhaps the most important one 
from a practical point of view. The other defect which we 
wish to emphasize is more of a theoretical nature. The parts 
of a triangle (its sides and angles) are usually given as 
numbers (so many feet, so many degrees). It seems natural, 
therefore, to suppose that there must exist some arithmetical 
process for the solution of triangles. 

Trigonometry enables us to find the unknown parts of a 
triangle by arithmetical processes. 

This statement must not be regarded as a complete defi- 
nition of trigonometry. We shall see later that the solution 
of triangles by arithmetical methods constitutes only a part, 
although an important part, of trigonometry. 

EXERCISE IV 

1. What are the principal sources of inaccuracy in solving a triangle 
by the graphic method? 

2. We wish to construct an isosceles triangle, given the angle at the 
vertex and the altitude. Suppose that the error made in constructing 
the angle is 5'. Will the effect of this error on the base of the triangle, 
as obtained from the construction, be different for different altitudes? 
Will it be greater or less for the higher triangles ? 

3. If you wish to find a point as the intersection of two straight lines, 
are you likely to obtain a more accurate result if the two lines are at 
right angles, or if they are nearly parallel ? 

4. A side and two adjacent angles of a triangle are gi\ r en, and its 
other parts are to be found graphically. Is such a graphic solution likely 
to be accurate if both of the angles are very small ? If the sum of the 
given angles is very close to 180° ? 

5. Formulate, in your own words, the distinction between an arith- 
metical and a graphical process. 



CHAPTER II 

THE TRIGONOMETRIC FUNCTIONS OF ACUTE ANGLES 

6. The necessity of introducing new ideas. At the close 
of Chapter I we formulated the problem of devising arith- 
metical methods for the solution of triangles. But, if we 
think of the many shapes which a triangle may assume, this 
problem appears to be a most formidable one. We shall, 
therefore, for the present, confine our attention to the com- 
paratively simple case of a right triangle. We have good 
reason to suppose that, if we succeed in solving our problem 
for all right triangles, we shall be able to deal 
later with the general case also, since every 
triangle may be decomposed into two right 
triangles. 

Let us then consider a triangle ABC, right- 
angled at (Fig. 5), and let us denote the 
sides opposite the angles A, B, C by a, 6, c, re- 
spectively. We are already acquainted with 
two relations which will assist us in the solution of our 
problem, namely : 

(1) A + B=90° 
and the theorem of Pythagoras, 

(2) a 2 + b 2 = c 2 . 

The first of these equations enables us to calculate one of the 
acute angles if the other one is given. The second provides 
a method for calculating any one of the three sides if the 
other two are given. 

But we have nothing as yet which will enable us to find 
the angle A if two of the sides (say a and b} are given, 
although the triangle is clearly determined by these sides 
and might be constructed by geometry. The equations (1) 

11 




12 FUNCTIONS OF ACUTE ANGLES 

and (2), unaided by other relations, are obviously inadequate 
for this purpose, since (1) is a relation between the angles 
A and B alone, while (2) involves only the sides of the tri- 
angle. Now, clearly, a statement which is only concerned 
with the angles of a triangle cannot convey any positive 
information about the sides, and vice versa. 

In order that we mag be able to solve a right triangle by 
arithmetical processes, there must, therefore, be added to equa- 
tions (1) and (2) certain other relations, in which the sides and 
angles shall not be separated, but in which they shall occur 
simultaneously. 

The discovery of such relations and their adequate formulation 
is the foundation upon which all of trigonometry must finally 
rest. 

A simple illustration will make clear the nature of these new relations. 

The numerical measure for the steepness of an inclined plane, say a 

mountain road, may be given in two ways. We may say that the road 

makes a certain angle A (say 5°) with the horizontal plane, or we may 

say that it rises a certain number of feet (say 87.5 feet) in a horizontal 

distance of 1000 feet. 

87 5 
The quotient, — — , is technically known as the slope, grtcde, or gradient 

of the road. The gradient is clearly connected with the angle A in such a 

way that, if A should vary, to every value 
of the angle there corresponds a definite 
value of the gradient and vice versa. Thus, 
if AB (Fig. 6) represents the road, and if 
a and b are both expressed in feet, the gra- 
dient of AB is equal to -. The value of this quotient changes with the 

b 
angle A, so that for different angles we find different values for the 
gradient. 

The general question before us may be formulated as fol- 
lows. If the acute angle A of the right triangle AB O be- 
comes larger or smaller, what effect will such a change have 
upon the sides ? And, conversely, if the sides of a triangle 
change, what effect will this have on the angles ? 

Now, since similar triangles have their corresponding 
angles equal, it is clear that there are some changes in the 




DEFINITIONS OF FUNCTIONS 13 

lengths of the sides which produce no change in the angles. 
In fact, if all of the sides of a triangle are magnified in the 
same ratio, the angles are not changed 
at all. 

In the two right triangles ABC 
and A'B'C (Fig. 7) we have 

a a' b b' a 



b b' 



= — , etc., 




if the ano'le at A' is equal to that at „ _ 

° 1 Fig. 7 

A. But if the angle A' is different 

from A, the ratios — , — , -— , etc., will not be equal to -, -, -, 
e g V c c b 

etc., respectively. For if they were, corresponding pairs of 

sides of the two triangles would have the same ratio, the 

triangles would be similar, and, contrary to our hypothesis, 

angle A! would have to be equal to angle A. 

Consequently, while the lengths of the individual sides of 

a right triangle have nothing to do with the size of its angles, 

the ratios of these lengths are connected with the magnitude 

of the angles in a very intimate fashion. In fact, so close is 

this relation that the values of the ratios -, -, -, etc., may be 

c c b 

determined (by construction) as soon as the acute angle A 
has been chosen, and conversely ; if one of these ratios is 
given, we can find (by construction) one, and only one, cor- 
responding acute angle A. 



7. Definitions of the trigonometric functions of an acute 

angle. We have seen that the values of the ratios -, -, etc., 

c c 

of the sides of a right triangle are closely bound up with the 
magnitude of the angle A. If the angle changes, each of 
these ratios changes and vice versa. 

Now, one variable quantity is called a function of another, if 
they are so related that any change in the latter produces a 
corresponding change in the former. 



14 FUNCTIONS OF ACUTE ANGLES 

Consequently, each of the six ratios -, -, -, -, -, -, deter- 

c c b a a b 

mined by the sides of the right triangle A B C, is a function 
of the angle A, because any change in A produces a corre- 
sponding change in each of those ratios. Each of these 
functions has received a name and a symbol. The reason 
for choosing these names will appear later (see Arts. 10 and 
70), and cannot be discussed with profit at the present 

moment. 

We proceed to give the formal definitions of 

the six trigonometric functions of an acute 

angle A. 




Construct any right triangle (cf. the Fig.), one 

of whose acute angles is equal to the given angle 

A.* Of the two legs of this right triangle, that 

one which passes through the vertex of the angle 

A is said to be adjacent to A. The other leg is said to be 

opposite to A, and the third side of the triangle is called its 

hypotenuse. 

The sine of A is the ratio of the opposite side to the 
hypotenuse. 

The cosine of A is the ratio of the adjacent side to the 
hypotenuse. 

The tangent of A is the ratio of the opposite side to the 
adjacent side. 

The cotangent of A is the ratio of the adjacent side to the 
opposite side. 

The secant of A is the ratio of the hypotenuse to the ad- 
jacent side. 

The cosecant of A is the ratio of the hypotenuse to the 
opposite side. 



* We have seen in Art. 6 that the size of this right triangle is absolutely of 
no consequence, since any two triangles of this kind are similar, so that the 
corresponding ratios for the two triangles will be equal. 



DEFINITIONS OF FUNCTIONS 15 



In symbols we may write these definitions as follows : 

a) 



4 <*> A ^ * 4 a 

sin A = , cos A = - , tan A = T , 
c c b 

4 c . c , . b 

csc A - — , sec A = = , cot ^L = — • 
a b a 

These symbols are written abbreviations of the names of the functions. 
In speaking, the symbols are pronounced as though the name of the func- 
tion had been written out in full. Thus, tan .1 is pronounced tangent 
of A or tangent A : esc .4 is pronounced cosecant of A or cosecant A, etc. 

In defining the trigonometric functions, we made use 
of the numerical measures of certain line-segments, namely, of 
the sides of a right triangle. Now, the numerical measure 
of a line-segment changes if the unit of measurement is 
changed. One might, therefore, expect the values of the 
trigonometric functions of an acute angle to change with 
every change of the unit of length. But this is not the case. 
Owing to the fact that only ratios of these line-segments 
appear in equations (1), the functions sin A, cos A, etc., are 
found to have the same value whether the line-segments a, 6, 
c are measured in feet, inches, or in terms of any other unit 
of length. 

Consider, for instance, a right triangle for which a — 3 feet, b = 4 
feet, c = 5 feet. According to (1) we have sin A = f, cos A = f , etc. 
Let us now introduce the inch as unit of length instead of the foot. The 
numerical measures of the sides of the triangle will now be a = 36, 
b — 48, c = 60. According to (1) we shall now find sin .4 = §{j, 
cos .4 = f|, etc. But f§ = f, |f = f, etc., so that we obtain precisely 
the same values for sin .4, cos A, etc., whether the sides of the triangle 
be expressed in feet or inches. 

If the trigonometric functions were concrete numbers, that 
is, if they were the numerical measures of some kind of con- 
crete quantity such as a length, an area, or a volume, their 
values would change every time that a change is made from 
one unit of length to another. We have just seen that this 
is not the case. Therefore, the trigonometric functions are 
pure or abstract numbers.* 

* This same fact is sometimes (somewhat inadequately) expressed by the 
statement that the trigonometric functions are ratios. 



16 FUNCTIONS OF ACUTE ANGLES 

8. The practical need of tables giving the values of the 
trigonometric functions. The trigonometric functions just 
defined will enable us to find the unknown parts of a right 
triangle when certain parts are given, provided only that we 
can devise a practical method for actually obtaining the 
numerical values of these functions for any given acute 
angle. Now, the values of the functions have been calcu- 
lated by mathematicians and the results have been collected 
in tabular form for convenient use. From the practical point 
of view, therefore, it only remains to explain the arrange- 
ment and the use of the tables. 

To the more scientifically inclined student, however, the 
question will immediately suggest itself as to how these use- 
ful tables were actually computed. We shall reserve the 
answer to this question for the second part of the book. 
The following examples, however, will show how a table of 
trigonometric functions may be prepared by the graphic 
method, provided that no very high degree of accuracy be 
required. 

EXERCISE V 

1. Find the functions of 40° by the graphic method. 
Solution. With the help of a protractor construct the angle PAQ 
(Fig. 8) equal to 40°. Choose any point i? on AQ and drop a perpendic- 
ular BC from B to A P. Measure the distances 
BC, AC, smdAB. Then will 

(1) sin 40°=— , cos 40°=— . 

v J AB AB 

P Although the point B might be chosen anywhere 

on AQ, it will be especially convenient to make AB 

equal either to one unit, ten units, or one hundred 

units. For, as equations (1) show, we have to divide by AB, and if 

AB is equal to 1, 10, or 100, we avoid the long division which would 

otherwise be necessary. 

Let us, therefore, make AB = 10 centimeters. We should then find 

by measurement, ^ D a a A ^ n „ 

J ' CB = 6.4 cm., AC = 7.7 cm. 

According to (1), therefore, 

sin 40° =M = 0.64, cos 40° = — = 0.77. 
10 ' 10 




VALUES OF FUNCTIONS 17 

Further we find 

tan 40° = — = 0.83, cot 40° = — = 1.20, 
7.7 6.4 

sec 40° = — = 1.30, esc 40° =— = 1.56. 
7.7 6.4 

We should obtain a more accurate result for tan 40°, and more con- 
veniently, if we were to use another triangle, making this time AC = 10 
cm. Measurement would then give BC = 8.4 cm., and 

tan 40°=— =0.84. 
10 

2. Fiud the functions of the following angles by the graphic method : 

(a) 10°. (b) 15°. (c) 20°. (d) 70°. 
Construct carefully each of the following right triangles, measure the 
angles, and find the six functions of each acute angle. 

3. a = 3, b = 4, c = 5. 

4. a = 5, 6 = 12, c = 13. 

5. (7 = 8, 5 = 15, c = 17. 



6. Construct and measure an acute angle whose sine is equal to 



7. Construct and measure an acute angle whose tangent is equal to f. 

8. Construct and measure an acute angle whose cosine is equal to f . 

9. Can you think of two right triangles (Fig. 7) with different 
angles .4 and A', for which the sides a and a' are nevertheless equal? 

10. Can you conceive of two right triangles (Fig. 7) with different 

angles A and A', for which the ratios - and — are nevertheless equal? 

b b' 

11. Why, then, do we speak of the ratio - as a function of A ? Why 

b 
do we not introduce a or ab as a function of A ? 

12. Assuming that we have access to a table of the values of the 
trigonometric functions, show how we might solve each of the following 
problems. To find the remaining parts of a right triangle when the 
following parts are given. 

I. o, b. C = 90°. III. a, c, C = 90°. 

II. a, A,C= 90°. IV. c, A, C = 90°. 

13. Show that neither the sine nor the cosine of an acute angle can 
ever be greater than unity. 

14. Show that the tangent of an acute angle may have any positive 
value whatever. Similarly for the cotangent. 

15. What restrictions, if any, are there on the values which the secant 
and cosecant of an acute angle may assume ? 



18 FUNCTIONS OF ACUTE ANGLES 

9. Relations between the six trigonometric functions of an 
acute angle. The preceding discussion suffices to indicate 
the importance of constructing a table of the values of the 
trigonometric functions. The task of computing these tables 
may be abbreviated very considerably by noting that the six 
functions are not independent of each other. In fact, we 
have (Fig. 9) 



(1) 




A <* 

smA = -, 

C 


A V 

csc -4. = -, 
a 


A h 

cos A = -, 
c 


sec A = -, 




tan .A = -, 




cot A = -, 
a 



so that we obtain at once the relations 

esc A = , secJ. = -, cot J. 



sin A cos A tan A 

sinJ. = -, cosJ. = -, tan^l = 



csc^.' seed.' cot A^ 

or, 

(2) sin ^1 csc ^4 = 1, cos^sed=l, tan ^ cot ^i = 1.* 

But, two numbers whose product is equal to unity are called 
reciprocals of each other. Therefore, equations (2) are 
equivalent to the following statement : 

The sine and cosecant, the cosine and secant, and finally the 
tangent and cotangent, of an acute angle are reciprocals. 

Clearly, knowledge of this fact reduces greatly the labor of computing 
tables of the functions. For, having found the values of the three 
functions, sine, cosine, and tangent, the values of the remaining three can 
be obtained from these by computing their reciprocals. 

But there are other relations besides (2) which enable us 
to reduce still further the labor involved in constructing a 
table of the trigonometric functions. We have 

tanA = f. 
o 

* Note that sin A esc A is written for sin A x esc A just as ab is written for 
axb. 



RELATIONS BETWEEN FUNCTIONS 19 

If we divide both numerator and denominator of the fraction 

- by c, we find 

b tonA = $/°. 

b/c 

But we have, by definition (cf. equations (1)) 

Q> A f> A 

- = sin A, ■- = cos A, 

c c 

and therefore 

(3) tan .4 = 5*1*4. 
v J cos A 

Of course, since cot A is the reciprocal of tan A, we also have 

(4) cot A =^4. 
v J sin A 

The relations (2), (3), (4) enable us to calculate the values 
of all six functions when the sine and cosine are known. 
But it actually suffices to know the sine. For, if we divide 
both members of the familiar equation 

(5) a 2 + b 2 = <? 
by c 2 , we find 



But, by definition, we have 

& • a b A 

— = &mA, -=cos-A, • 

c c 

so that (5) becomes 

(6) sin 2 ^4 + cos 2 ^ = 1, 

where sin 2 A and cos 2 A have been written, as is customary, 
for (sin A) 2 and (cos A) 2 , respectively. 

It follows from relations (2), (3), (4), and (6) that, if we 
know the value of a single one of the six trigonometric 
functions of an acute angle, the values of the remaining five 
may be computed. The detailed proof of this statement will 
be left to the student in some of the examples given below. 



20 FUNCTIONS OF ACUTE ANGLES 

EXERCISE VI 

In each of the following twelve examples, the value of one function of 
the acute angle A is given. Find the values of the remaining functions. 

1. sin ^4 = f. 4. cot^4=2. 7. sin A = x. 10. 'cot ^4= £. 

2. cos^= T 5 3. 5. sec^4 = ] r f. 8. cosA=x. 11. sec-A=x. 

3. tanJ.=l. 6. csc A = ty. 9. t&nA=x. 12. cscA=x. 

13. Construct an isosceles right triangle and make use of this figure 
for the purpose of computing the functions of 45°. 

14. Divide an equilateral triangle into two right triangles by dropping 
a perpendicular from one of its vertices to the opposite side. Make use 
of this figure for the purpose of computing the functions of 30° and 60°. 

15. Collect in a table the results of Exs. 13 and 14. 

16. Prove the formula sec 2 A = l + tan 2 A. 

17. Prove the formula esc 2 A — 1 + cot 2 A. 

18. Prove that the sine, tangent, and secant of an angle increase when 
the angle grows from 0° to 90°. 

19. Prove that the cosine, cotangent, and cosecant of an angle decrease 
when the angle grows from 0° to 90°. 

20. Prove that tan A < 1, cot A > 1 if A < 45°, and that tan A > 1, 
cot ,4 < 1 if .4 > 45°. • 

21. Show that each of the six functions may be expressed either as a 
product or as a quotient of two of the others. 

10. Relations between the functions of complementary 
angles. As a result of the relations discussed in the preced- 
ing article, the problem of computing the values of the six 
trigonometric functions for every angle between 0° and 90° 
has been reduced to that of computing these values for a 
single one of these functions. But we may reduce the prob- 
lem still further by observing the relations between the 
functions of the two acute angles of the same right triangle. 
We have (cf. Fig. 10) 

sini = -, cosi? = -, 
c c 

b . ^ b 




a b • t> b 
cos A = -, sm B = -, 

o. o. 



COMPLEMENTARY ANGLES 21 



tan^. = -, 




cot B = -, 




cotA = -, 

a 


tanJB = -, 


A C 

sec A = -, 




csc i> = -, 
6 


A C 

csc A — -, 
a 
and therefore 


sec B = -, 


^v sin ^4. = cos i?, tan A = 
cos^A = sini?, cot A = 


: cot 5, sec ^4 = csc i?, 


: tan j5, csc A = sec J5. 



Since the angles A and i? are complementary, we may 
write these equations as follows : 

sin (90° - A) = cos A, cot (90° - A) = tan A 
(2) cos (90° -A) = sin 4, sec (90° - A) = csc .4, 

tan (90° -A) = cotA, csc (90° - A) = sec A 

An easy way to remember these formulae is as follows: 
Let the six functions be grouped into three pairs : sine and 
cosine, tangent and cotangent, secant and cosecant. Let us 
speak of either function of one of these pairs as the cof unction 
of the other. Then, the six formulae (2) are all included in 
the following statement. 

Any trigonometric function of the complement of an angle A 
is equal to the cof unction of A. 

It is apparent that this theorem will enable us to find the 
trigonometric functions of any acute angle greater than 45°, 
if we know the functions of all angles less than 45°. Thus, 
for instance, tan 75° is equal to cot 15°, sin 82° is equal to 
cos 8°, etc. As a consequence of this fact it is possible to 
reduce the space occupied by the tables of the functions 
to exactly half of what would otherwise be necessary. 

The relation between the functions of complementary angles is also 
important in another respect. It is this relation which has given rise to 
the words cosine, cotangent, and cosecant. The cosine is the sine of the 
complement. At a time when Latin was still the universal language of 
the scientific world, the cosine was therefore called complement sinus. 



22 



FUNCTIONS OF ACUTE ANGLES 



This was later (in the seventeenth century) contracted to cosinus. The 
words cotangent and cosecant originated in the same manner. 



EXERCISE VII 

1. Express as functions of the complementary angles 

sin 37°, cos 62°, tan 13°, cot 75, sec 12° 15', esc 55° 37'. 

2. If the table of values of the functions is so arranged as to give only 
the functions of angles less than 45°, how may we obtain the values of 

sin 57°, cos 63° 15', tan 75° 12', cot 67° 18' ? 

3. What acute angle is that whose sine is equal to the sine of its 
complement? 

4. Find an acute angle for which tan A = cot (45° -{-A). 

Hint. Substitute for tan^l its equal cot (90° — ^4) and note that two 
acute angles with the same cotangent are equal to each other. 

5. Find an acute angle for which sin 2 A = cos (45° — A). 

6. Find an acute angle for which cot 3 A = tan 2 A. 

7. Find an acute angle for which cos A = sin 6 A . 

8. Find an acute angle for which sec 2 A = esc 7 A. 

11. The values of the functions of o°, 30 , 45 , 6o°, 90 . 

While we have decided to postpone the general question of 
the arithmetical calculation of the trigonometric functions, 
we have already performed this calculation for a- few special 
angles, viz. : 30°, 45°, 60° (cf. Exs. 13, 14 of Exercise VI). 
The figures there suggested, and the results are as follows : 



0) 



sin 45°= cos 45° = ^= 

tan 45°= cot 45° = 1, 
sec45°=csc45° = V2. 



V2 



= hvw, 



(2) 



sin 30° = cos 60° 



2_^ 
c 



1 

2' 




cos 30°= sin 60° = JV3, 

tan 30°= cot 60° = -k±- = JL = I V3, 



l C V3 



cot 30°= tan 60°= V3, 
sec 30°= esc 60° 

esc 30°= sec 60° 



1.V3 



V3 



V3 3 



1 


3 


6/ 


\ 

a? V 


/GOV 


<s \ 



6= He C 
Fig. 12 




VALUES FOR SPECIAL ANGLES 23 

An acute angle of a right triangle can never be 0° or 90°, 
so that the definitions of Art. 7 are not applicable to such 
angles. The acute angle A may, however, approach either 
0° or 90° as a limit, and if it does, its functions in some cases 
approach definite finite limits. By sin 0°, 
cos 0°, sin 90°, cos 90°, etc., we mean sucji 
limits whenever they exist. 

In Fig. 13, let the angle A = PAQ be 
thought of as decreasing toward the limit a ^ b 
zero as a result of the rotation of the side FlG 13 

AQ around A as a center, while the side 
AP remains fixed. Through any point C of AP draw CR 
perpendicular to AP, and denote by B the intersection of 
OR with the rotating line AQ. 

As the angle A approaches zero, the side a approaches 
zero and c approaches b. Consequently 

sin A = - approaches - or 0, cos A = - approaches - or 1, 

CO CO 

tan A = - approaches - or 0, sec A = - approaches - or 1. 
bo ob 

In this sense we may say that 

sin 0° = 0, cos 0° = 1, tan 0° = 0, sec 0° = 1. 

The function cot A = - has no limit when A approaches zero. 

For, the ratio - grows larger and larger as A approaches 

zero, since b remains fixed while a grows smaller and smaller. 
Clearly, this ratio may be made larger than any number 
however great, by choosing A and, hence, a small enough. 
This is expressed in symbolic language as follows : 



or in words : The cotanyent of an acute angle increases without 
bound when the angle approaches zero- as a limit. 

A similar argument holds for esc A = -, since c approaches 



24 FUNCTIONS OF ACUTE ANGLES 

b and a approaches zero. Hence, in symbolic language, we 

have • 

r8 v f sin 0° = 0, tan 0° = 0, sec 0° = 1, 

^ } I cos 0° = 1, cot 0° = oo, esc 0° = oo. 

By a similar argument the student may deduce the follow- 
ing results : 

( 4 . J sin 90° = 1, tan 90° = oo, sec 90° = oo , 

^ ) I cos 90° = 0, cot 90° = 0, esc 90° = 1, 

the exact meaning of each of which should be expressed in 
words as in the cases which have just been treated in extenso. 



CHAPTER III 

SOLUTION OF RIGHT TRIANGLES BY NATURAL 
FUNCTIONS 



12. Arrangement and use of the table of natural functions. 

The numerical values of the trigonometric functions are 
usually called the natural functions to distinguish them from 
the logarithms of these functions which we shall study later. 

Table * V gives the numerical values of the sine, cosine, 
tangent, and cotangent to four decimal places for every tenth 
of a degree from 0° to 90°. The values of the secant and 
cosecant are omitted because they are not used very fre- 
quently. They may of course be calculated, whenever 
necessary, by the formulae 

1 , 1 



sec A = 



esc A 



(Art. 9) 



cos A sin A 

The following is a sample portion of the table. Only this 
part of the table will be required for the following illustra- 
tive examples. 



AXGT.K 


N Sw 


* 


N Tan 


d 


N Cot 


d 


X Cos 


(7 




35°.0 

.1 


0.5730 
0.5750 


14 
14 


0.7002 
0.7028 


26 
26 


1.4281 
1.4229 


52 
53 


0.8192 
0.8181 


11 
10 


55°.0 

54°. 9 


.2 


0.5764 


15 


0.7054 


26 


1.4176 


52 


0.8171 


10 


.8 


.3 


0.5779 


14 


0.7080 


?' 


1.4124 


53 


0.8161 


10 


.7 


.4 


. 
















.6 


.5 


















.5 


N Cos 


d 


N Cot 


d 


N Tax 


d 


X Bis 


d 


Angle 



* See Logarithmic and Trigonometric Tables, compiled by E. J. Wilczynski 
and H. E. Slaught. 

25 



26 SOLUTION OF RIGHT TRIANGLES 

Problem 1. Find the functions of 35°.2. 

Solution. In the left-hand column find 35°. 2. The four numbers 
which are printed in the horizontal row to the right of 35°.2 are, from left 
to right, the sine, tangent, cotangent, and cosine of 35°.2, as indicated by 
the name printed at the head of each of these columns. Therefore 

sin 35°.2 = 0.5764, tan 35°.2 = 0.7054, cot 35°.2 = 1.4176, 

cos 35°.2 = 0.8171. 

Problem 2. Find the functions of 54°.8. 

Solution. In the right-hand column find 54°. 8. The four numbers 
which are printed in the horizontal row to the left of 54°. 8 are, from right 
to left, the sine, tangent, cotangent, and cosine of 54°.8, as indicated by 
the name printed at the foot of each of these columns. Therefore 

sin 54°.8 = 0.8171, tan 54°.8 = 1.4176, cot 54°.8 = 0.7054, 

cos 54°.8 = 0.5764. 

Thus every number of the table does double duty. For 
example, 0.5764 is both the sine of 35°. 2 and the cosine of 
54°.8, as it should be. (See Art. 10, equations (2).) 

Angles less than 45° are given in the left-hand column of 
the table, and the names of the corresponding functions are 
found at the top of the page. Angles greater than 45° are 
given in the right-hand column with the names of the func- 
tions at the bottom of the page. 

The table gives the values of the functions only for every 
tenth of a degree. If the given angle contains fractional 
parts of this unit, its functions cannot be read directly from 
the table. In such cases we make use of the process of in- 
terpolation, the nature of which will become apparent from 
the following examples. 

Problem 3. Find the sine of 35°.17. 

Solution. This angle lies between 35°. 1 and 35°.2. More precisely, it 
lies T 7 o of the way from the former toward the latter. We conclude that 
its sine will be ^ of the way from 

sin35°.l = 0.5750 toward sin 35°.2 = 0.5764. 

But the difference d between these last two numbers is 0.0014, seven 
tenths of which is equal to 0.0010 (reduced to four decimal places). 
Therefore 

sin 35°.17 = 0.5750 + 0.0010 = 0.5760. 



BY NATURAL FUNCTIONS 27 

Problem 4. Find the cotangent of 35°.17. 
Solution. From the table we find 

cot 35°. 1 = 1.4229 
cot35°.2 = 1.4176 
d = cot 35°.2 - cot 35°. 1 = — 0.0053 (tabular difference) 

We must add T V of d to cot 35°. 1. But ^d = - 0.0037. 
Therefore cot 35U7 = 1.4192. 

We observe that in problem 3 the correction was positive, 
while in problem 4 it was negative. 

If ice always interpolate from the smaller toward the larger 
angle, the correction will be positive in the case of sine and tan- 
gent, negative in the case of cosine and cotangent. For, the 
former two functions increase with the angle, while the latter 
two decrease. 

There will never be any serious danger of giving the wrong sign to 
the correction, if we cultivate the habit of running through the numbers 
of the table near the place we are using, so as to see in which direction 
they are growing. 

EXERCISE VIII 

1. Find all of the functions of 15°.3, 28°.7, 63°.4, 82°.l. 

2. Find sin 37°.24, cos 62°.19, tan 53°.42, cot 27°.13. 

3. Formulate in words the principle upon which the method of inter- 
polation is based. Is this principle absolutely exact, or is it in the 
nature of an approximation? 

4. What are the numbers in the four narrow columns of the table 
headed rf, and what purpose do they serve? 

5. In the arrangement of the table as explained, what use has been 
made of the results of Art. 10 ? 

We have shown how to find the functions when the angle 
is given. It remains to show how to find an angle when one 
of its functions is known. The general method will be ap- 
parent from the following examples. 

Problem 5. The tangent of an unknown acute angle A is equal to 
0.7046. Find the angle A. 

Solution. In the specimen table on page 25 we observe that the num- 
ber 0.7046 does not occur in the tangent column. However, we find there 



28 SOLUTION OF RIGHT TRIANGLES 

the two numbers 0.7028 and 0.7054 between which 0.7046 lies. Thus we 

have tan 35°.l = 0.7028, 

tan A = 0.7046, 
tan 35°.2 = 0.7054. 

Between tan35°.l and tan A, the difference is 0.0018. 
Between tan 35°.l and tan 35°.2, the difference is 0.0026. 
Therefore, tan A is £f of the way from tan 35°.l toward tan 35°. 2, and 
consequently A = 35<u + ^ of Qne tenth q{ ft degre ^ 

or 

A = 35°.l + 0°.07 = 35°.17, 
reducing to the nearest hundredth of a degree. 

Problem 6. Find the acute angle A whose cosine is 0.5772. 

Solution. We have 

cos54 .7 = 0.5779, 
cos A = 0.5772, 
cos 54°.8 = 0.5764, 
whence 

cos A - cos 54°.7 = - 0.0007, 
cos 54°.S - cos 54°. 7 = - 0.0015. 

Therefore, cos A is T % of the way from cos 54°.7 toward cos 54°.8. Hence 

A = 54°.7 + T \ of one tenth of a degree 
or A = 54°.7 + 0°.05 = 54°.75. 

With a little practice, the student will soon become suf- 
ficiently expert in the process of interpolation to enable him 
to perform this operation mentally. He should train him- 
self with that end in view. 

If we wish to find, by means of our table, the natural 
functions of an angle which is expressed in degrees, minutes, 
and seconds, the minutes and seconds should first be converted 
into decimal parts of a degree. This may easily be done, 
remembering that 1' = g 1 ^ of a degree and 1" = ^ of a minute. 
Table VII may be used to save time in making this trans- 
formation. 

EXERCISE IX 

1. Find the values of 

sin 18° 12', cos 67° 23', tan 58° 34', cot 64° 16'. 

2. Find the acute angles for which 

sin A = 0.5673, tan C = 1.7328, 

cos B = 0.2791, cot D = 0.8924. 



BY NATURAL FUNCTIONS 29 

13. Solution of right triangles by means of the table of 
natural functions. A right triangle is determined by any 
two of its parts (not counting the right angle) provided that 
at least one of these parts is a side. The relations, which 
we have found between the angles and sides, enable us to 
compute the remaining parts of a right triangle when any 
two such parts are given. In order to make sure of this 
fact, let us see what cases may present themselves. 

If both of the given parts are sides, there are two possi- 
bilities. Either the two given sides include the right angle 
(Case 1), or else one of the given sides is the hypotenuse 
(Case -)• If one of the given parts is a side and one is an 
angle, we have again two possibilities, viz. : given the hy- 
potenuse and one acute angle (Case 3) ; or, given one leg 
and one acute angle (Case 4). We proceed to discuss these 
cases in order. 

Cask 1. Given the two sides of the triangle which include 
the right angle {the tivo legs'). 

In our previous notation this means that a and b are given. 
In this case we may first compute 

tan A = % 

o 

and then find A from the table. We may then find c from 

, _ a ^ 

sin A cos A 

in two ways (affording a check), and B from 

B = 90° - A. 

Case 2. G-iven one leg and the hypotenuse. 

We may denote the given leg by a. Then a and c are 
given. The solution is accomplished by means of the for- 
mulae 

>mA = -. b = aG0tA = ccQaA, B = W)°-A. 
c 



30 SOLUTION OF RIGHT TRIANGLES 

Case 3, Given the hypotenuse and one acute angle. 

We may denote the given acute angle by A. Then A and 
c are given. The solution is given by the equations 

a = c sin A, b = c cos A, B = 90° — A. 

Case 4. Given one leg and one acute angle. 

Since the knowledge of one acute angle implies that of the 
other, we may assume that a and A are the given parts. To 
find the remaining parts, we use the formulae 

b = acotA, c = — — = — — t , B = 90°-A. 
sin A cos A 

Our discussion has shown that the methods at our dis- 
posal suffice to find the remaining parts of a right triangle 
when two independent parts of the triangle (not counting 
the right angle) are given. To establish this fact was the 
purpose of the above classification. It is not necessary, nor 
even desirable, when solving a numerical problem of this 
sort, to find out first under what case it falls. In practice it 
is better not to refer to this classification at all, but to pick 
out and solve those among the four equations 

(1) sin A = -, cos A = -, tan A = -, A + B = 90° 

ceo 

which contain only one unknown quantity each. The remain- 
ing equations may then, in most cases, serve as a check. 

A more complete check is given by the equation 

(2) a 2 + b 2 = c 2 . 

In order to avoid the inconvenience of forming the quan- 
tities a 2 , 5 2 , e 2 by actual multiplication, we have supplied a 
table of squares (Table VI). The arrangement and use of 
this table will be apparent from the following examples. 

Example 1. Find the squares of 0.324 and of 3.24. 

Solution. In the left-hand column of Table VI we find 0.32. In the 
same horizontal row with this number, and in the column headed 4, we 
find 0.1050. Therefore 

(0.324) 2 = 0.1050, (3.24) 2 = 10.50. 



BY NATURAL FUNCTIONS 31 

Example 2. Find the squares of*0.3243, of 3.243, and of 3:2.43. 

Solution. From the table we find 

(0.324) 8 = 0.1050, (0.325) 2 = 0.1056. 

The difference between the two squares is 0.0006. The number 0.3243 
is three tenths of the way from 0.3:21 toward 0.325. Therefore, iis square 
will be three tenths of the way from 0.1050 toward 0.1056. That is 

(0.3243)* = 0.1050 + T % of 0.0006 = 0.1050 + 0.0002 = 0.1052, 
aiul (3.243) a = 10.52, (32.43) 2 = 1052. 

Example 3. Find the square root of 0.5520. 

Solution. We find from the table that this number is the square of 

0.743. Therefore 

V0.552O = 0.743. 

Example 4. Find the square root of 0.5525. 

Solution. The table gives 

VO5520 = 0.743, V0.5535 = 0.744. 

Therefore, by interpolation 

V05525 = 0.743 + & of 0.001 = 0.743 + 0.0003 = 0.7433. 

Iii engineering practice, equation (2) is used very exten- 
sively in connection with such tables of squares, not merely 
for checking, but for the purpose of performing the original 
calculation. The tables of Inskip and Smoley are particu- 
larly convenient for this purpose, if the distances are ex- 
pressed in feet, inches, and thirty-seconds of an inch. 

The following examples illustrate the methods for solving 
right triangles. 

Example 5. Tn a right triangle, right angled at C, given a = 3.479, 
b = 2.321. Compute the remaining parts of the triangle. 

Solution. We find first 

tan A =^ = M[9 = 1.4989. 
b 2.321 

The table of tangents gives 

A = 56°.2 + %i of 0°.l = 56°.30, 

reducing to the nearest hundredth of a degree, as usual. The tables of 
sines and cosines give 

sin A = 0.8320, cos A = 0.5548. 



32 SOLUTION OF RIGHT TRIANGLES 

We compute next 

3 ' 479 =4.181, 



sin A 0.8320 

and use the equation c cos A = b as a check. We find 

c cos A = 2.320, b = 2.321. 

The two members of the" check equation do not agree exactly, but we 
have no right to expect absolute agreement. All of the numbers used in 
the calculation are merely approximations, giving us the values of the 
functions to the nearest unit of the fourth decimal place. In combining 
several such approximate numbers, the error may occasionally exceed 
two or three units of the last decimal place. 
Finally we find 

B = 90° - A = 33°.70. 

We may exhibit this solution more compactly as follows. The figures 
in parentheses indicate the order in which the various results are ob- 
tained. 

Formulae. tan A =-, c = — — , B = 90° - A. 

b sin A 

Check, c cos A = b. 

a = 3.479 (1) sin A = 0.8320 (5) c = 4.181 (7) 

b = 2.321 (2) cos A = 0.5546 (6) c cos A = 2.320 (8) Check. 

tan ,4 = 1.4989 (3) A = 56°.30 (4) B = 33°.70 (9). 



Given 



Example 2. In a right triangle, right angled at C, given c = 5.783, 
A = 42°.39. Compute the remaining parts of the triangle. 

Solution. Formulae, a = c sin A, b = c cos A, B — 90° — A. 
Check. a 2 + 2 _ c 2 # 

Given ! c = 5 - 783 W a = 3 - 899 ^ 

U =42°.39 (2) 6 = 4.271 (7) 

B = 47°.61 (3) a 2 = 15.20 (8) 

sin A = 0.6742 (4) ft 2 = 18.24 (9) 

cos A = 0.7386 (5) a 2 + 6 2 = 33.44 (10) 1 Check 

c 2 = 33.44 (11) } 

Remark. The quantities a 2 , & 2 , c 2 required for the check were obtained 
from the table of squares. When no such table is available, it is usually 
desirable to write the check equation in the form 

a 2 = c 2_ b 2 = ( C _ & )( c + & ) ? 

since this form of the equation reduces by one the number of multiplica- 
tions required. 



BY NATURAL FUNCTIONS 33 

In this example, the check computation would then yield 
a-= 15.202, c-b = 1.512, 

c + b = 10.054, 
C '2 _ &a _ 15.202 Check.- 

EXERCISE X 

In each of the following right triangles, right angled at C, two parts 
are given. Compute the remaining parts and check. Also check by 
means of a graphic solution to provide against gross errors. 

1. a = 27, A = 25°.l. 5. c = (104.5, .4 = 47°.53. 

2. a = 34.5, c = 52°.8. 6. a = 8.695, £ = 7.321. 

3. a = 2.7S1. & = 3.056. • 7. b = 62.78, c = 81.39. 

4. 6 = 87.95, ,1 = 55°.36. 8. B= 29°.58, c = 2354. 

9. A gravel roof slopes three fourths of an inch per horizontal foot. 
What angle does it make with a horizontal plane? 

10. The pitch of a gable roof is the quotient obtained by dividing the 
height of the ridge-pole above the garret floor by the width of the garret. 
What is the pitch of a gable roof covering a garret 38 feet wide, if the 
ridge-pole is 15 feet above the garret floor, and what angle does the roof 
make with a horizontal plane ? 

11. At a time when the sun was 55° above the horizon, the shadow of 
a certain building was found to be 112 feet long. How high is the 
building ? 

12. The side of a regular decagon is 3.471 feet. Find the radii of the 
inscribed and circumscribed circles. 

13. The side of a regular polygon of n sides is equal to a. Find 
formulae for the radii of the inscribed and circumscribed circles. 



CHAPTER IV 

DISCUSSION OF SOME DEVICES FOR REDUCING THE 
LABOR INVOLVED IN NUMERICAL COMPUTATIONS 

14. The number of decimal places. As we attempted to 
point out in Chapter I, every number obtained as a result of 
measurement is really an approximation. If we measure the 
distance between two dots on our drawing board, by means 
of a carefully constructed scale which reads to the fiftieth 
part of an inch, we may still estimate half of this smallest 
scale unit with the naked eye. Let us assume that the 
divisions of the scale are reliable, that the ruler is very 
nearly straight, that the dots are very small, and that we are 
using the greatest of care in our measurement. We may 
then concede that the result of such a measurement (say 5.34 
inches) is accurate to the nearest T ^ of an inch. This 
means that no number, with two figures to the right of the 
decimal point, is as close to the true value as 5.34. It means 
that 5.33 is certainly too small and that 5.35 is certainly too 
large. It does not mean that the true value is exactly 5.34 
inches, but that the true value lies between 5.335 and 5.345 
inches. 

When we record the result of such a measurement, the 
number of decimal places which we write (two in this 
example) is an indication of the degree of precision 
which we claim for the result. In this connection, let us 
note emphatically that a zero, when obtained as the last digit 
of the measure of a quantity, should never be suppressed. 
Suppose, for instance, that in the above example we had 
obtained 5.30 inches as the result of our measurement. 
This means that we are certain that the true value of the 
distance lies somewhere between 5.295 and 5.305 inches. 

34 



ACCURACY OF RESULTS 35 

If we were to record this result as 5.3 inches, suppressing 
the final zero, we should be giving the erroneous impression 
that we had measured the distance only to the nearest tenth 
of an inch and that it might have any value between 5.25 
and 5.35 inches. 

Thus, the number of decimal places, which we use in record- 
ing the result of a measurement, is an indication of the degree 
of precision which we attribute to this result. 

This being so, we are guilty of negligence whenever we 
express such a result by a decimal with fewer places than we 
are able to guarantee. For we are thus throwing away 
knowledge which we have actually had in our possession. 
But it would be dishonest to express our result with more 
decimal places than we can guarantee. For we should then 
be tending to mislead others into thinking our measurements 
more accurate than they really are. 

We may estimate the degree of precision of a number, as we 
have just done, on an absolute scale. But clearly it will 
usually be more reasonable to adopt as a measure of pre- 
cision the ratio of the Wk probable error " of the measurement 
to the total magnitude of the quantity measured. If we do 
this, an error of one foot in a thousand is to be regarded 
as of no greater importance than an error of yoVo" °^ an 
inch in an inch. In either case we may say that it is an 
error of -^ of one per cent. 

Whenever we properly record the result of a measurement 
by a number consisting of four digits, no matter where the 
decimal point may be placed, this means, in the light of our 
preceding discussion, that the error of the last digit is guar- 
anteed to be less than half a unit of the last decimal place. 
Xow, the smallest number expressed by four digits is 1000. 
Let us suppose that the exact value of our unknown quan- 
tity is x = 1000 units, but that as a result of our measure- 
ment we have found x = 999.5 units. Then our error is 
2 oVo' or 2V °f one P er cent i °f the total magnitude measured. 
The largest number expressible by four digits is 9999. If 



36 NUMERICAL COMPUTATIONS 

the exact value of x is 9999 and our error of measurement is 
half a unit, the error will be to the total magnitude measured 
as J is to 9999, or as 1 is to 19998. It will be an error of 
about 2^-q of one per cent of the whole. Thus, four digits 
are certainly sufficient to express the result of a measurement 
if its degree of accuracy does not exceed ^ of one per cent. 
Now,' most of the ordinary operations of surveying fall well 
within this limit, so that four decimal places are usually 
sufficient to express the results obtained by the surveyor. 

15. The accuracy of a sum, difference, product, or quotient 
of two numbers obtained by measurement. It is clear that 
the sum or difference of two numbers can have no greater 
precision than the less accurate of the two numbers. Con- 
sequently, it is useless and misleading to retain more decimal 
places in one term of a sum or difference than in the others. 

In forming a product we are apt to do a great deal of use- 
less work if we fail to remember that the factors, and there- 
fore the product, are mere approximations. Suppose we 
have measured the sides a and b of a rectangular field to the 
nearest hundredth of a foot and have found a = 35.67 ft., 
6 = 86.72 ft. To find the area of the field we form the 
ftR 79 product ab. The ordinary method (shown in 
' R n the margin) gives the result 3093.3024 square 
feet. But if we allow the result to stand in 



~™o^ this form, we shall exhibit either our ignorance 
o2032 i • j. p i " • • t- 

AQonft or our desire to create a false impression, tor 

9 ir this would seem to indicate a result precise to 

q nq q o (l 9i the nearest y"o o o"o °^ a S( l uare foot, whereas it is 
uncertain by more than a whole square foot as 
we shall now show. 

In fact, the equations a =35.67 ft., 5 = 86.72 ft. merely 
mean that a and b are between the limits 

35.665 < a < 35.675, 86.715 < b < 86.725 
respectively, so that the area must be between the limits 

86.715 x 35.665 < ab < 86.725 x 35.675 
or 3092.685475 < ab < 3093.919375. 



LABOR-SAVING DEVICES 37 

The uncertainty in the value of ab is therefore so great 
that no digit to the right of the decimal point has any real 
significance. Even the last digit preceding the decimal 
point is not certain, so that we are even slightly overstating 
our accuracy, if we write simply 

ab = 3093 square feet. 

Thus, we see that the product of two approximate four- 
place numbers is to be regarded as accurate to no more than 
four places. Consequently, it is a waste of time and labor to 
actually work out all of the partial products in the multipli- 
cation. We might abbreviate the process as follows : 

86.72 or better, by 86.72 

35.67 writing the 35.67 



6. 


more important 


2602. 


52. 


partial products 


434. 


434. 


first ; 


52. 


2602. 




6. 



3094. 3094. 

The fact that we find 3094, instead of 3093, need not dis- 
turb us, since we have seen from our above considerations 
that the last digit is actually uncertain to the extent of one 
unit. 

This process, which is sometimes called abbreviated multi- 
plication, is obviously preferable to the ordinary method, 
since it does away with the labor of first finding numbers 
which must afterwards be discarded. 

Similar remarks may be made for division. More generally, 
whenever we are dealing with numbers whose first four digits 
only can be regarded as certain, it is wise to abbreviate all 
of our calculations correspondingly. It should be remarked, 
however, that, in certain exceptional cases, very exact results 
may be obtained from inaccurate data and vice versa. But 
this is not the place for a discussion of such cases. 

16. Labor-saving devices. We have seen that extensive 
calculations, even with four-place numbers, are apt to be 



38 NUMERICAL COMPUTATIONS 

troublesome and laborious. In many problems it is neces- 
sary to use five- six- or seven-place numbers. In such cases, 
the amount of labor required becomes excessive, even if we 
make use of the abbreviated method of multiplication and 
division. 

A very important aid in performing such calculations is 
furnished by certain tables, such as tables of squares (of 
which Table VI is an example), tables of cubes, and recip- 
rocals of numbers, etc. Crelle's Tables, which are merely 
a systematically arranged and extensive set of multiplication 
tables, are particularly valuable. 

A second great aid to numerical calculation is furnished 
by calculating machines, which are now being used very 
extensively in commercial as well as scientific work. The 
ordinary cash register is one of the simplest of these 
machines. 

Graphic methods for solving numerical problems consti- 
tute a third great class of aids to calculation. These meth- 
ods have, in recent times, been modified and extended, so 
as to become capable of greater accuracy, and for many 
problems no other method of solution is known. The 
slide rule, which may be classed either with the graphic 
methods or among the calculating machines, has become 
so popular among engineers as to exclude, in their work, 
almost all other methods of calculation. (See Arts. 29 
and 30.) 

But the most important of all of these labor-saving de- 
vices, without which the slide rule and many similar con- 
trivances cannot even be thoroughly understood, is the 
method of calculation by logarithms. 

17. Definition of logarithms. It is apparent, from what 
has been said, that the real cause of the laboriousness of 
extensive calculations lies in the operations of multiplication 
and division. Addition and subtraction, even of numbers 
with many places, are comparatively easy. It was this fact 
which caused John Napier (1550-1617) and Jobst Burgi 



DEFINITION OF LOGARITHMS 39 

(1552-1632)* to consider the possibility of devising a 
method, by means of winch addition and subtraction might 
be made to do the work of multiplication and division. The 
method which they invented for this purpose is essentially 
equivalent to the one which we shall now explain, though 
very different from it in form. We must remember that the 
notations of modern algebra, to which we are accustomed 
and which are of the greatest assistance to us in our mathe- 
matical arguments, are the results of a century-long process 
of development. This process was far from complete in 
Napier's and Biirgi's time. The greatness of the achieve- 
ment of these men can only be properly appreciated when 
judged from the standpoint of the mathematical knowledge 
of those days.f 

From our present point of view the possibility of reduc- 
ing the operations of multiplication to that of addition is 
an immediate consequence of a familiar fact of algebra. 
This fact, embodied in the formula 

a x a y = a x+y , 

states that the product of two powers of the same base is 
itself a power of that base, whose exponent is equal to the 
sum of the exponents of the two original powers. We may 
express this fact by the statement that to the multiplication 
of two powers of the same base corresponds the addition of 
their exponents. By this simple remark, multiplication is 
actually converted into addition for all such numbers as are 
known powers of a common base. 

We shall assume that the fixed number a, the base of our 
system, is positive and different from unity. If the expo- 
nent x is a positive integer, there will be comparatively few 

* Napier was of Scotch and Burgi of Swiss nationality. Biirgi's discovery 
of logarithms was unquestionably independent of Napier's and was made at 
about the same time. But Napier's book " Mirifici Logarithmorum canonis 
description containing an account of his method was published in 1614, six years 
earlier than Biirgi's " Arithmetische und Geometrische Progress- Tabula." 

t For an excellent account of the history of logarithms see Cajori in The 
American Mathematical Monthly, Vol. 20 (1913). 



40 NUMERICAL COMPUTATIONS 

numbers which can be regarded as powers of a. Suppose, 
for example, that a = 2. Then 2, 4, 8, 16, 32, etc., are 
integral powers of 2, but 1, 3, 5, 6, 7, 9, 10, etc., are not. 
But, from our previous study of algebra, we are acquainted 
with the fact that the symbol a x may be defined, not merely 
for the case when the exponent is a positive integer, but also 
when the exponent is any positive or negative rational num- 

ber of the form ± - (j? and q being integers), or zero. 

These definitions are as follows : 

If x is a positive integer (x = £>), 

I. a x = a v = a • a • a ••• (a product of p factors each equal 
to a). 

If x is a positive rational fraction lx = ~\ 

II. a x - a p/Q = Va p = QVa) p , 
where ~Va means the positive ^th root of a. 

If x is a negative rational fraction ( x = ), 

in. a * = «-*/* = JL= J_. 

Finally, if x = 0, 

IV. a* = a* = 1. 

Now, there is nothing remarkable about the fact that we 
have been able to define the symbol a x in all of these cases. 
We might have done that in many different ways. But it 
is remarkable that, if we adopt these particular definitions, 
the formulae 

(1) a x • av = a x+y , 

(2) - = a x ~y, 

(3) (a x y = a xy 

turn out to be true, not only when the exponents are posi- 
tive integers but in all of the four cases for which we have 



DEFINITION OF LOGARITHMS 41 

defined the symbol a x . That this should be so, is of course 
not merely a fortunate coincidence. It is due to the fact 
that the definitions II, III, IV were deliberately chosen in 
such a way as to insure the universal validity of formulae 
(1), (2), and (3). These formulas, which are collectively 
known as the three index laws, are fundamental for the fol- 
lowing discussion. 

Equations I, II, III, IV suffice to define the symbol a x 
whenever x is a rational number. Now, every irrational 
number can be approximated, as closely as we may desire, by 
means of a decimal fraction ; and this decimal fraction 
(which is a rational number) will take the place of the original 
irrational number in all numerical calculations. We may 
define a x , when x is irrational, as follows : 

Let x x be the closest approximation, to the irrational num- 
ber x, which is possible by means of a decimal fraction with 
only one figure to the right of the decimal point. Let x 2 be 
the closest approximation possible by means of a decimal 
fraction with only two figures to the right of the decimal 
point. Let x 3 , x# ••• x n , ••• be similar approximations with 
3, 4, ••• n figures after the decimal point. The sequence of 
rational numbers . _ 

has the irrational number x as a limit. Then a x is defined 
to be the limit of the second sequence of numbers 

a x i, a x *, a**, • • • a*», • • • .* 

As an example, consider a = 10, x =V2. We have 

x x = 1.4, x 2 = 1.41, x 3 = 1.414, x A = 1.4142, etc., 

«*. = 10H a" = 10 1 - 41 , a*3 =10!- 414 , a* = 10 1 - 4142 , etc. 

By 10^2 we mean the limit approached by the numbers of 
this second sequence. 

For practical purposes, however, V2 is replaced by one of 
the approximations 1.4, 1.41, 1.414, etc., namely, the first 
one which is sufficiently accurate for the particular problem 

* These limits exist, but it would carry us too far to prove this fact. 



42 NUMERICAL COMPUTATIONS 

under consideration ; and lO^ 2 " is replaced by the first one 
of the numbers 10 1 - 4 , 10 1 - 41 , etc., which is sufficiently close to 
the true value for the purposes of the problem in question. 

It may be shown that, if the above definition of a x for 
irrational values of x be adopted, the index laws will hold also 
for irrational exponents. 

We may now state, without formal proof, a theorem which 
is fundamental in the theory of logarithms, in so far as the 
very existence of logarithms depends upon it. This theorem 
is as follows : 

If a is a positive number different from unity, there exists 

one and only one exponent x (positive, negative, or zero), such 

that Y xr 

a x = N, 

where N is any positive number. 

Although we state this theorem without proof, the student 
may easily convince himself of its great plausibility by a 
process which, if carried out to its logical conclusion, would 
constitute a proof. Suppose, for instance, that a = 2 and 
that N= 1000. We have 

2 6 = 32, 2 6 = 64, 2^ = 128, 2 8 = 256, 2 9 = 512, 2 10 = 1024. 
We conclude that the value of x for which 

2* = 1000 
must be between 9 and 10. Now 21/2= V2 = 1.4142 .... 
Therefore 

2 9.5 = 29.2i/2= 512 x 1.4142 = 724.1. 
But 2 io = 10 24 ? 

so that x must lie between 9.5 and 10. We may obtain 
closer and closer limits between which x must lie by con- 
tinuing this process, and thus ultimately show that there 
exists a number x (as a limit of a sequence), for which 
2 X = 1000. This argument at the same time indicates a 
process by means of which the exponent x may be calculated 
to any desired number of decimal places. 



PROPERTIES OF LOGARITHMS 43 

We are now ready to define a logarithm. 

The logarithm of any positive number N* with respect to the 

bate a, is the exponent of the power to which the base a must be 
raised in order to obtain the number N.* 

In other words, if a* = JV, 

we say that x (the exponent) is the logarithm of N with 
respect to the base a. In symbols we write this same state- 
ment as follows : x _ ] ()CT ^r 

Example. Since 5 3 = 125, we have log 5 125 = 3. 



EXERCISE XI 

1. What are the logarithms of 2, 4, 8, 16, 32, 64, 128 with respect to 
the base 2? Write out each of these results in symbols; thus, log 2 4 = 2. 

2. What are the logarithms of 3, 9, 27, 81, 243 with respect to the 
base 3? 

3. What are the logarithms of 10, 100, 1000, 10,000 with respect to 
the base 10? 

4. What are the logarithms of 3, 9, 27, 81, 243 with respect to the 
base 27? 

5. What are the logarithms of 1, |, $, -V, ^r, 2I1 W1 th respect to the 
base 3 ? 

6. What are the values of 2 X . 3 Z . 4 X , 10 z when x is equal to zero? 
What. then, is the logarithm of 1 with respect to each of the bases 2, 3, 
4. 10? 

7. What is the logarithm of 1 with respect to any base a? 

8. What is the logarithm of any number with respect to itself as 
base? 

9. Find, approximately to two decimal places, the number whose 
logarithm, with respect to the base 2, is equal to 1.5. 

18. The properties of logarithms. Those properties of 
logarithms which are of importance for the purposes of nu- 
merical calculation, are immediate consequences of the 
index laws and of the definition of logarithms. In fact, we 

*The word logarithm is derived from the Greek Ao Y o? or logos, meaning pro- 
portion or ratio, and dptO^ or arithmos, meaning number. The reason for 

rhonsing this name will be apparent from the theorem stated in Exercise XII, 
Ex.4. 



44 NUMERICAL COMPUTATIONS 

may write each of two positive numbers, M and JV, in the ' 
form 

(1) M = a\ JST= ay, 

so that, in accordance with the definition of logarithms, 

(2) x=log a M, y=log a K 

According to the first index law (Art. 17, equation (1)), 
the product of M and N is equal to 

MN— a x -a y = a x+ v, 

whence, by the definition of logarithms, 

log a (MW) = x + y = log a M+ log a N. 

The theorem expressed by this formula may obviously be 
extended to any number of factors. Hence, 

I. The logarithm of a product is equal to the sum of the 
logarithms of its factors. 

From (1) we obtain by division, 

N ' 

making use of the second index law (Art. 17, equation (2)). 
Therefore, by the definition of logarithms, 

loga [jy\ = x - y = log a M- log a iV, 

a result which may be formulated as follows : 

II. The logarithm of a quotient is equal to the logarithm 
of the dividend minus the logarithm of the divisor. 

The same fact may, of course, be stated in the equivalent 
form: the logarithm of a fraction is equal to the logarithm of the 
numerator minus the logarithm of the denominator. 

According to the third index law (Art. 17, equation (3)), 
we have (*■)*= a*» 



PROPERTIES OF LOGARITHMS 45 

Therefore, we find from (1) 

M p = <f'\ 

or, by the definition of logarithms, 

log m M*=px=p\og a M. 
Consequently, 

III. The logarithm of the p th power of a number M is ob- 
tained by multiplying the logarithm of 31 by p. 

Since the third index law is true whether p be an integer 
or a fraction, the last theorem has the following corollary. 

obtained bv putting p equal to - : 

n 

IV. The logarithm of the n'" root of a number M is obtained 
by dividing the logarithm of 31 by n. 

The content of the two equations 

a 1 = a, a = 1 

may be stated as follows : 

V. Tlie logarithm of any number, with respect to itself as 
base, is equal to unity. 

VI. TJie logarithm of unity, with respect to any base, is 
equal to zero. 

It is clear now how logarithms will serve to reduce the 
operations of multiplication and division to those of addition 
and subtraction. Suppose that we have at our disposal a 
table of logarithms. To multiply M by N we look up the 
logarithms of these numbers from the table and add them 
together. We then find the number 'whose logarithm is 
equal to this sum by again referring to the table ; this num- 
ber is the product MN. To divide M by N we proceed in 
the same way. except that in this case we form the difference 
log M — log N instead of the sum. 



46 NUMERICAL COMPUTATIONS 

EXERCISE XII 

1. If k>g 10 2 = 0.30103, log 10 3 = 0.47712, find log 10 12, logic (I), log 10 (¥), 
log io #6. 

2. Express, in terms of log B _p and log« q, the following quantities : 

l0ga(pY), l0g.(^), log^E 4 . 

3. Prove the truth of the following statement. If logio x is expressed 
as a decimal fraction (x being a positive number greater than unity), 
the logarithm of \0 k x (k being a positive integer) will differ from logi z 
only in its integral part. 

4. Prove the theorem : If the numbers a v a 2 , a 3 , ... are in geo- 
metrical progression, their logarithms are in arithmetical progression. 

5. Prove the equation 

log* x + V x2 " 1 = 2 log a (x + V^Tl). 
a; _ Vx 2 - 1 



CHAPTER V 

CALCULATION BY LOGARITHMS 

19. Common logarithms. With scarcely an exception, the 
civilized nations of all times have made use of the decimal 
Bystem for expressing numbers, both in the spoken and in the 
written language.* For this reason, the number 10 is es- 
pecially well adapted to serve as base for a system of 
logarithms. Logarithms with respect to the base 10 are 
usually known as common logarithms ; they are also some- 
times called Briggsian logarithms, in honor of Henry Briggs 
(1556-1630), f who constructed the first table of common 
logarithms. 

In this book we shall have very little occasion, hereafter, to 
speak of any except the common logarithms. We shall 
therefore agree to abbreviate the symbol log 10 iVto log iV, 
the base 10 being understood when no other base is men- 
tioned explicitly. 

The positive integral powers of 10, such as 10, 100, 1000, 
etc., the negative integral powers of 10, such as 0.1, 0.01, 
0.001, etc., and the zero power of 10, which is equal to 1, 
are the only numbers whose common logarithms are integers. 
The logarithms of all other numbers have an integral and a 
fractional part. 

Tlie fractional part of the logarithm is called the mantissa, 
while the integral part of the logarithm is known as its char- 
acteristic. 

* It is usually admitted that the predominance of the decimal system over all 
others is due to the fact that the normal human being has ten fingers. This 
opinion has certainly been generally held since the time of Aristotle. 

t Briggs was the first Savilian Professor of Geometry at Oxford. According 
to Ball (see Ball's Primer of the History of Mathematics) , Briggs was also the 
first to make systematic use of the decimal notation in working with fractions. 

47 



48 CALCULATION BY LOGARITHMS . 

20. Properties of the mantissa. We consider the mantissa 
and the characteristic separately because, in practice, the 
method for finding the characteristic of a logarithm is entirely 
different from that employed for finding its mantissa. The 
reason for this will appear from the following discussion. 

Let us consider an example. From the definition of a 
common logarithm, we know that 

(1) log ^10 = log 10* = 1 log 10 = l-l = 0.25000. 

Now it is not difficult to compute V10 by elementary 
methods. We may, for instance, first compute VlO to as 
many decimal places as we desire, and then extract the square 
root of the result. We find, in this way, to five decimal 
places, 

(2) ^10 = 1.75792 
or, if we combine (1) and (2), 

(3) log 1.75792 = 0.25000. 

From the theorem about the logarithm of a product, we 
conclude 
log 17.5792 = log (1.75792 x 10) = log 1.75792 + log 10 

= 0.25000 + 1 = 1.25000, 
log 175.792 = log (1.75792 x 100) = log 1.75792 + log 100 
= 0.25000+2 = 2.25000, 

We observe that the numbers 1.75792, 17.5792, 175.792, 
etc., contain the same succession of digits, and differ from 
each other only in the position of the decimal point. Their 
logarithms, on the other hand, whose values we have just 
calculated, differ from each other only in the value of the 
characteristic. 

Again, if we make use of the theorem about the logarithm 
of a quotient, we find from (3) 

log 0.175792 = log 1 - 75 J 92 = 0.25000 - 1, 
log 0.0175792 = log L ^ 92 = 0.25000 - 2, 



DETERMINATION OF THE CHARACTERISTIC 49 

Now the negative quantities, which appear on the right 
members of these equations, are not written in the form 
which we ordinarily use for negative quantities. Thus, for 
instance, we have found the value of log 0.0175792 to be 
0.25000 — 2, a result which we should ordinarily write in the 
form — 1.75000, to which it is obviously equal. If we agree 
to write every negative logarithm in this unusual form, 
as a difference between a positive proper fraction and an 
integer, thus making its fractional part positive, we gain the 
advantage that the mantissas will be the same for any two 
numbers which contain the same succession of digits, even if 
none of these digits appear to the left of the decimal point. 
We avoid, in this way, the necessity of using two different 
tables of mantissas, one for numbers greater, and one for 
numbers less, than unity. 

Let us recapitulate the result of our discussion in two 
formal statements. 

I. We agree to express the logarithm of any positive number 
N in such a form that its mantissa shall be positive. 

This can be done whether log N is positive or negative, 
that is, whether JV be greater or less than unity. In the 
latter case, the negativeness of log N is brought about en- 
tirely by means of the negative characteristic. 

As a consequence of this agreement, the following state- 
ment will be true in all cases. 

II. If two numbers contain the same succession of digits, that 
is, if they differ only in the position of the decimal point, their 
logarithms will have the same mantissa and will differ only in 
the value of the characteristic. 

It is for this reason that the tables give only the mantissas 
of the logarithms and that, in looking up the mantissas, we 
pay no attention to the position of the decimal point in the 
given number. 

21. Determination of the characteristic. The characteristic 
of a logarithm is easily determined by inspection. Its value 



50 CALCULATION BY LOGARITHMS 

depends merely on the position of the decimal point. Since 
we have 

10° = 1, 10 1 = 10, 10 2 = 100, 10 3 = 1000, etc., 
or 

log 1 = 0, log 10 = 1, log 100 = 2, log 1000 = 3, etc., 

we draw the following conclusions : 

If 1 < iV< 10, then < log iV< 1. .• . log N has the 
characteristic 0. 

If 10 < JST< 100, then 1< log 1ST < 2. . • . log JST has the 
characteristic 1. 

If 100 < N< 1000, then 2 < log JST< 3. . • . log JV has 
the characteristic 2. 



If 10*<iV r < 10 *+i, then k< log N<k + 1. .-. logi^has 
the characteristic k. 

We may formulate these results as follows : 

III. If k is a positive integer, and if the number N" lies be- 
tween 10* and 10* + 1 , the characteristic of log N is equal to k. 

Since such a number N has k + 1 digits to the left of the 
decimal point, we obtain the following rule : 

IV. If N is any number greater than 1, the characteristic of 
its logarithm is one less than the number of digits in its integral 
part. 

The student is advised to make but little use of this rule 
on account of its mechanical character. Statement III pro- 
vides a better method (less mechanical and easier to re- 
member) for determining the characteristic. 

It remains to show how to find the characteristic of log N 
when N< 1. 

If .1 < N< 1, then - 1< log N< 0. . • . log N has the 
characteristic — 1. 

If .01 < N< .1, then - 2 < log N< - 1. .-. log N has 
the characteristic — 2. 



USE OF THE TABLE OF LOGARITHMS 51 

If .001 <i\T< .01, then - 3< LogiV< -2. .-. log JSThaa 
the characteristic — 3. 



li WTi < * < W>' then - (*+ *)< l °s N< -*•• ••• lo s 

.V has the characteristic — (& + 1). 

ExamiuatioD of this table leads to the following two state- 
ments, either of which may be used to determine the char- 
acteristic of log iVwhen N< 1. 

If k is a positive integer, and if the number N lies between 

- - and — - — , the characteristic of log N is — (k -f- 1). 
10* 10* - 1 

7f JV" i's fass fcAtfw 1, and is expressed as a decimal fraction 
having k zeros between the decimal point and the first significant 
figure, then the characteristic of the logarithm of Nis — (k + 1). 
In one of our illustrations we had found 

log 0.0175792 = 0.25000 - 2. 
We must never write this in the form 

log 0.0175792= -2.25000, 
since only the characteristic is negative and not the fractional part. 
Some computers use the notation 

log 0.0175792 = 2.25000 ; 

but for most purposes it is preferable to write 

log 0.0175792 = 8.25000 - 10, 
and similarly 

log 0.175792 = 9.25000 - 10. 

In other words, in actual practice, we write a positive char- 
acteristic 10 — k in place of the negative characteristic — k, and 
then subtract 10 from the whole logarithm. 

22. Arrangement and use of the table of logarithms. We 

have already mentioned the fact that the table of logarithms 
gives only the mantissas. The characteristics must be sup- 
plied by the computer by the methods of Art 21. 

The table which we shall ordinarily use (Table 1) gives 
the mantissa, for every number from 1 to 9999. to five decimal 
places. 



52 



CALCULATION BY LOGARITHMS 



In order to explain the arrangement of this table, we shall 
reprint a small portion of it, and solve a number of typical 
examples, chosen in such a way as to require only this part 
of the table for their solution. 



N 





l 


2 


3 


4 


5 


6 


7 


8 


9 


PP 




34242 

439 
635 
830 


262 
459 
655 
850 


282 
479 
674 
869 


301 

498 
694 
889 


321 
518 
713 
908 


341 

537 
733 

928 


361 

557 
753 
947 


380 

577 
772 
967 


400 
596 

792 
986 


420 

616 

811 

*005 




19 


20 


220 
221 
222 
223 


1 
2 
3 
4 
5 
6 
7 
8 
9 


1.9 

3.8 

5.7 

7.6 

9.5 

11.4 

13.3 

15.2 

17.1 


2.0 

4.0 

6.0 

8.0 

10.0 

12.0 

14.0 

16.0 

18.0 



Problem 1. Find the logarithm of 221.4. 

Solution. To find the mantissa we ignore the decimal point. We 
read down the left-hand column of the table (headed N) until we find 
the first three digits of our number, viz., 221. The numbers printed in 
the same horizontal row with 221 are, in order, the mantissas of the 

logarithms of 2210, 2211, 2212, , 2219, as indicated by the 

number at the head of each of the next ten columns. To save space, 
however, the first two digits of the mantissa are never printed more than 
once in each row. In our case we find the mantissa, from the column 
headed 4, to be .34518. Since 221.4 is between 100 = 10 2 and 1000 =10 8 , 
the characteristic is 2. Therefore 

log 221.4 = 2.34518. 

Problem 2. Find log 22.39. 

Solution. Looking for the mantissa as before, we find *005. The 
asterisk indicates that the first two digits of the mantissa are not 34, as 
one might suppose, but 35. The reason for this appears clearly from the 
J^able. Therefore 

log 22.39 = 1.35005. 

If the number JV contains more than four digits, its loga- 
rithm cannot be read directly from the table. But it may 
be found by interpolation. We illustrate this process by an 
example. 



USE OF THE TABLE OF LOGARITHMS 53 

Problem 3. Find log 221.73. 

Solution. From the table we find, supplying the characteristics our- 
selves, 

log221.70 = 2.34772 

log 221.80= 2.34792 

Tabular difference = 0.00020 = 20 units of the fifth decimal place. 

Since 221.73 is ^ of the way from 221.70 toward 221.80, we add ft of 
the tabular difference to log 221.70. Therefore 

log 221.73 = 2.34772 + & of 0.00020, 
or 

log 221.73 = 2.34772 + 0.00006 = 2.34778. 

The auxiliary tables in the margin, headed P P (abbre- 
viation for proportional parts), facilitate the process of in- 
terpolation. 

Thus, in problem 3, we refer to the auxiliary table with 20 (the 
tabular difference) at its head. In the third row we find r 3 of 20 or 6.0. 

It remains to show how to find the number when its loga- 
rithm is given. 

Problem 4. Given log A 7 = 9.34857 -10. Find the value of N to 
five significant figures. 

Solution. The characteristic of logiV is 9—10 or — 1. Therefore, 
the number N must be between 10 _1 = 0.1 and 10"= 1. Consequently, 
the decimal point will precede the first significant figure of iV. 

The mantissa 34857 does not occur in the table, but it falls between 
the two tabular mantissas 34850 and 34869. 
Thus we have : 

9.34S50 -10= log 0.22310 (from the table), 
9.34857 -10 = log N, 

9.34869 - 10 = log 0.22320 (from the table), 
so that N lies between 0.22310 and 0.22320. 

We observe that logiV lies T 7 ¥ of the way from log 0.22310 toward 
log 0.22320. Therefore, N lies ft of the way from 0.22310 toward 0.22320. 
That is. 

2V= 0.22310 + T ^ of 10 units of the fifth decimal place. 
But 

T Sj of 10 units = {$ units = 3)] units = 1 units. 

Therefore 

X= 0.22310 + 0.00004 = 0.2231 1. 



54 CALCULATION BY LOGARITHMS 

Also in this inverse problem (to find the number when its 
logarithm is given) interpolation is aided by the auxiliary 
tables in the margin. 

Thus, in problem 4, the tabular difference is 19. The difference be- 
tween log N and the smaller of the two tabular logarithms, between 
which log N lies, is 7. The auxiliary table with 19 at its head, shows 
that, among the tenths of 19, that one, which comes closest to the value 
7, is the fourth. Consequently, N is T 4 - of the way from 0.22310 toward 
0.22320. Therefore, up to five decimal places, N= 0.22310 + 0.00001 
= 0.22314. 

23. Cologarithms. Since we obtain the same result whether 
we divide iVby M, or multiply iVby — , we may, in a loga- 
rithmic calculation, add the logarithm of —-instead of subtract- 

ing log M. The logarithm of ~~ is called the cologarithm of M. 
Therefore 

colog M= log — = log 1 — log M = — log M, 

since log 1 is equal to zero. 

Cologarithms, like logarithms, are written with positive 
mantissas. Consequently, the cologarithm of a number is 
most easily found by subtracting its logarithm from zero, 
written in the form 10.00000 — 10, as in the following 
example. 

Problem 5. Find the cologarithm of 221.73. 

Solution. 

10.00000 - 10 
log 221.73 = 2.34778 
colog 221.73 = 7.65222 - 10 

It is easy to perform this operation of subtraction from 
10.00000 — 10 mentally. There is no gain, however, from the 
use of cologarithms when we are dealing with a quotient of 
only two numbers. A real advantage is gained by the in- 
troduction of cologarithms, when more than two logarithms 
are to be combined by addition and subtraction. For the 
logarithms which are to be subtracted we then substitute 



EXTRACTION OF ROOTS 55 

cologarithms, enabling" ns to complete the operation by a 

single addition. 

It often happens, just as in the case of forming a cologa- 
rithm, that we wish to subtract a logarithm from another 
smaller one. In all such cases we change the form of the 
minuend by adding and subtracting 10, or some convenient 
multiple of 10, as in the following example. 

32 34 

Problkm 6. Compute —£ — . 
472.3 

Solution. We find from Table I, 

log32.34 = 1.50974, 
log 47-2.3 = 2.(57 422. 

Iu order to subtract the latter logarithm from the former, we write 
log 32. 34= 11.50974-10,* 
log 472.3 = 2.67422 

logi 3 Mi = 8.83552 -10 
8 472.3 

Hence, from the table, P^ = 0.068473. 
472.3 

24. Extraction of roots by means of logarithms. Since 

log <J~x = log x Vp = - log ^ 

p 

it is easy to extract roots of any order by means of logarithms. 
If the characteristic of log x is not negative, no further re- 
mark is necessary. If log x is negative, we proceed as in the 
following example : 

Problem 7. Compute by logarithms : V.53760. V.53760, and V.53760. 
Solution. Log 0.5:5700 = 9.73046-10. 



log V.53760 = \ log 0.53760 = \ (19.73046 - 20) = 9.86523 - 10. 
log V^3760 = i log 0.53760= \ (29.73016-30) =9.91015-10. 



log V.53760 = l log 0.53760 = \ (49.73046 - 50) = 9.94609 - 10. 
Therefore, from Tab i. 

V.53760 = .73322. v^.53760 = .81312, V.53760 = .88326. 



• A computer with some experience will refrain from actually writing the 
logarithm in the form 1 1 ..~>0074 — 10. It is easy for him to carry out the calcula- 
tion at though it were so written. 



56 CALCULATION BY LOGARITHMS 

25. Logarithmic calculations which involve negative numbers. 

We have only defined the logarithms of positive numbers. 
But this suffices for our purposes. Clearly, when we com- 
pute a product or quotient, its numerical value may be found 
first, without paying any attention to the signs of the various 
factors. Afterwards, the proper sign ( + or — ) may be pre- 
fixed to the result according as there is an even or an odd 
number of negative factors. 

The easiest way to keep a count of the negative factors is 
to use the method, introduced by Gauss,* of writing the 
letter n immediately after a logarithm which corresponds 
to a negative number. In forming a sum or difference of 
logarithms, we write an n after the result only if an odd 
number of the separate logarithms is affected by an n. 

Example. If N = - 221.73, we write 

log N = 2.34778 n. 

EXERCISE XIII 

■ 1. Making use of the tables, find log 3726, log 67.43, log 729800, 
log 0.3896, log 0.008527. 

2. Making use of the tables, find log 32653, log 76.431, log 879450, 
log 0.045723, log 0.0059426. 

3. By means of the tables, find the numbers whose logarithms are 
3.84522, 1.68079, 8.89064 - 10, 7.12548 - 10, 2.27068. 

4. By means of the tables, find the numbers whose logarithms are 
3.89067, 9.24110 -10, 1.52195. 

5. Given a = 3.1572, b = 7.2916, c = 45.731. Compute by logarithms 
the values of ab, be, ca. 

ab 



6. With the same values of a, b, c compute 

c 

3 / 27, 

7. With the same values of a, b, c compute -v 



8. Compute the volume of a hemispherical dome if its diameter is 
150.32 feet. (Volume of a sphere of radius r is f 7rr 3 .) 

*C. F. Gauss (1777-1855) was without question one of the greatest and most 
versatile mathematicians of all times. He was director of the observatory and 
professor of astronomy at Gottingen from 1807 to the end of his life. 



LOGARITHMS OF THE TRIGONOMETRIC FUNCTIONS 57 

9. If a sum ol money P (the principal) is earning interesl at the 

rate of r ', a year, and if the interesl is added to the principal at the end 

of each year, show that the amount, at the cud of n years,, will be 

A = P 1 1 + JL) n . 

V 100/ 

In this ease the interest is said to be compounded annually. 

10. Find the amount on $157.38 for 7 years at 3£% compound 
interest. 

11. I low much money must I put into the bank at 3% compound 
interest, so that the amount may be $ 500 at the end of five years? 

26. The logarithms of the trigonometric functions. In solv- 
ing right triangles by means of logarithms, we frequently 
have to find the logarithm of a trigonometric function of an 
angle. It would be very burdensome if we had to look up 
first, in the table of natural functions, the value of the func- 
tion and then, from the table of mantissas, find its loga- 
rithm. In order to avoid this complication, there has been 
constructed an additional table which enables us to find 
directly the values of the logarithms of the sine, cosine, tan- 
gent, and cotangent for every angle between 0° and 90°. 
These quantities are denoted by the symbols log sin or 
L. Sin, log cos or L. Cos, etc., and are pronounced log sine, 
log cosine, log tangent, and log cotangent. 

In the tables which accompany this book, Table II gives 
the values of the logarithms of the trigonometric functions 
directly, to five decimal places, for every minute of arc. If 
the angle contains fractional parts of a minute, we obtain its 
functions from the table by interpolation. 

The arrangement of this table resembles that of the table 
of natural functions so closely, that it is unnecessary to de- 
scribe it in detail. It should be noted, however, that in 
this table the characteristics of the logarithms are also given. 
But since the natural sines and cosines of all acute angles, 
and the tangents of all angles less than 45°, are proper frac- 
tions, these characteristics are negative and have been ex- 
pressed in the form 9 — 10, 8 — 10, etc. TJie continually re- 
rurrin<i — 10 has not ht>en printed, and should be supplied by 



58 



CALCULATION BY LOGARITHMS 



the computer. It is understood, once for all, that 10 is to be 
subtracted from all of the logarithms in the first, second, and 
fourth columns of the table, while the logarithms printed in 
the third column are provided with their correct character- 
istics and require no such modification. 

The process of interpolation may be applied to the table of 
logarithms of the trigonometric functions in the same way 
as to the table of natural functions or to the table of loga- 
rithms of numbers. 

The following examples are intended to illustrate the ap- 
plication of Table II. 

Example 1. Find log sin, log cos, log tan, log cot of 41° 15' 35". 
Solution. For convenience in interpolation convert 35" into decimal 
parts of a minute. Then 41° 15' 35" = 41° 15' 58. 

We find, from the table, the following material: 

41° 



1 


L Sin 


D 


L Tan 


o r> 


L Cot 


LCos 


D 




P P 


15 
16 


9.81911 
9.81926 


15 


9.94299 
9.94324 


25 


0.05701 
0.05676 


9.87613 
9.87601 


12 


45 
44 


2 
3 
4 
5 
6 
7 
8 
9 


15 

1.5 

3.0 

4.5 

6.0 

7.5 

9.0 

10.5 

12.0 

13.5 




LCos 


D 


L Cot 


O D 


L Tan 


LSin 


D 


i 





48° 
We conclude : 

log sin 41° 15'.58 = 9.81911 + .58 of 15 units of the 5th decimal place, 
log tan 41° 15'.58 = 9.94299 + .58 of 25 units of the 5th decimal place, 
log cot41°15'.58 = 0.05701 - .58 of 25 units of the 5th decimal place, 
log cos 41° 15'.58 = 9.87613 - .58 of 12 units of the 5th decimal place. 

We may use the marginal tables of proportional parts to complete the 
interpolation. Thus, the table headed 15, shows that .5 of 15 is 7.5 and 



THE ACCURACY OF FIVE-PLACE TABLES 59 

.08 of l") is 1.2, and consequently .58 of L5 is 8.7 or !» units of the fifth 
decimal place. Therefore 

log sin I L5'.58 = 9.81920 1<>. 
In the same way we find 

log tan 11' 15'.58 = 9.94314 - 10. Log cot 11 15'.58 0.05686, 
log cos 11 15'.58 = 9.87606 - 10. 

Example 2. Find the Logarithms o\ the functions of 48 1 1'.42. 

§ 'inn. This an git 1 is the complement of thatof Example 1. Hence 
each oi iis functions is equal to the corresponding cofunction of 41° 15'. 58, 
and the values obtained are the same as in Example 1 with the name of 
each function changed to the corresponding cofunction. 

Just as in the table of natural functions, these values, for angles 
greater than l.V. may he obtained directly from the table by reading the 
degrees of the angle at the foot of the page 1 , the minutes in the right- 
hand column, and the name of the function at the foot of each of the 
four columns. We find, in this way. 

log .sin is l L'.42 = 9.87606 - 10, log cot 48°44'.42 = 9.94314 - 10, 

log tan *8°44'.42 = 0.05686, log cos 48°44'.42 = 9.81919 - 10. 

Example 3. Given log tan A = 0.53219. Find A. 
Solution. The given logarithm does not appear anywhere in the col- 
umn at the foot of which is printed the name L Tan. But we do find 
in this column log tan 73° 38 =0.53212, 

log tan 73° 39 = 0.53259 . 
Tabular difference for 1' = 0.00047. 
The given value of log tan A is ? 7 ~, or ^, of the way from the first 
toward the second of these tabular logarithms. Therefore 

.4 = 73° 38M5 

27. The accuracy of five-place tables. As we have said 
repeatedly, the number of decimal places used in stating the 
result of a measurement is to be regarded as an indication of 
its degree of precision. We shall ordinarily wish to make 
all calculations, based upon such measurements, with a suffi- 
cient number of decimal places to avoid introducing inaccu- 
racies which might have an appreciable influence upon the 
results, i.e. an influence comparable with that produced by 
the unavoidable errors of observation. 

Five-place tables arc quite accurate enough to satisfy this 
condition in almost all problems of engineering and natural 



60 CALCULATION BY LOGARITHMS 

science. In fact, in most problems of this kind, the distances 
are not measured so accurately as to exclude an error of one 
twentieth of one per cent of their value, and the angles are 
read to the nearest minute only. But five-place tables are 
far more accurate than this. In fact, a distance expressed 
by a five-place number presupposes an accuracy of at least 
2 Jo" %' an( * an angle may usually be determined from five- 
place logarithms of its functions with an error of not more 
than 2 or 3 seconds of arc. 

We must not forget, however, that the degree of accuracy 
of a table is not the same in all parts of the table, and that 
we must use our judgment in the selection of the formulae 
which we wish to use in solving a problem. 

28. The trigonometric functions of angles near o° or go°. 
Thus, for instance, if we wish to determine an angle for which 
log cos A = 9.99998 — 10, our table cannot furnish an accu- 
rate result. We find, by referring to the table, that A may 
have any value between 0° 29' and 0° 36'. 

A small angle cannot be determined, ivith any degree of 
accuracy, from the value of its cosine. 

In the same way, we see that an angle very close to 90° 
cannot be determined accurately from the value of its sine. 

In most cases we shall be able to modify the formula, 
which we are using, in such a way as to avoid this difficulty. 
If, for instance, the angle A (known to be very small) is to 
be determined from the value of its cosine, we shall seek 
some other formula as a solution of the same problem by 
means of which the angle A can be determined from the 
value of its sine or tangent. The problem then reduces to 
that of finding a small angle when its sine or tangent is 
given. If we again refer to our table, we find that this 
problem also gives rise to a difficulty. The method of inter- . 
polation, which we ordinarily use, becomes both cumbersome 
and inexact in the case of such small angles, because the 
tabular differences are very large and change very rapidly 
from one place in the table to another. 



THE LOGARITHMIC OB GUNTER SCALE 61 

In order to meet this difficulty, we have provided a sepa- 
rate table (Table III), giving- the values of the logarithmic 
functions for every second of arc from 0° 0' to 0° 3' and from 
89° 57' to 00°, and for every ten seconds from d to 2° and 
from 88° to 90°. 

Another method of meeting this difficulty, preferable in 
some respects, will be explained in the second part of this 
book (Art. 85); it involves the auxiliaries S and T (Table 
IV). 

EXERCISE XIV 

1. Find log - sin 15 3 6', log cos 35° 13', log tan 57° 28', log cot 76° 44'. 

2. Find log sip 18° 23'.35, log tan 41° 46'.27, log cos 64° 17' 43", log cot 
26 c 12' 38". 

3. Find the angles for which 

log sin A = 9.42553 - 10, log cos B = 0.60618 - 10 
log cot C = 0.6S407 = 10, log tan J) = 0.14193. 

4. Find the angles for which 

log cot A = 0.11157, log tan B = 9.75465 - 10, 

log cos C = 9.6S334 - 10, log sin D = 9.56652 - 10. 

5. Find log sin 0° 2' 15". log tan 1° 10' 22". 

6. Find the angles for which 

log sin A = 5.83170 - 10, log tan B = 8.32313 - 10. 

7. Find, by logarithms, the angle A, if tan .4 = a/b and a = 1.2291, 

b = 14.950. 

29. The logarithmic or Gunter scale. No graphical process 
is more familiar than the addition and subtraction of line- 
segments, and this process may evidently be used as a sub- 
stitute for addition and subtraction of numbers. Since addi- 
tion of logarithms corresponds to multiplication of numbers, 
we may find the logarithm of a product graphically by add- 
ing line-segments, whose lengths are equal to the logarithms 
of the factors. 

In order to do this, we must have some means for actually 
finding a line-segment whose length shall be equal to the 
logarithm of a given number. 



62 CALCULATION BY LOGARITHMS 

Let us take a line-segment of convenient length, say 10 
centimeters, as unit of length. In terms of this unit, the 
whole distance (10 centimeters = 100 millimeters) repre- 
sents log 10, since the logarithm of 10 is equal to unity. 
If we count all distances from the left-hand end of the line, 
we may label the right-hand end 10 to indicate that this 
distance represents log 10. The left-hand end will then be 
labeled 1, because log 1=0. 

From the table of logarithms we have, to two decimal places, 

log 1=0.00, log 2 =0.30, log3 = 0.48, log! = 0.60, 

(1) log 5 = 0.70, log6 = 0.78, log7 = 0.85, log8 = 0.90, 

log 9 = 0.95, log 10 = 1.00. 
We mark the points on our line-segment whose distances 
from the left-hand end, measured in terms of the whole line 
as unit, are in order equal to log 2, log 3, log 4, ... log 9, 
and label them 2, 3, 4, ... 9, respectively. If the whole 
line-segment is 10 centimeters long, these points will, on 
account of (1), be at distances 30, 48, 60, 70, 78, 85, 90, 95 
millimeters, respectively, from the left-hand end of the line- 
segment (cf. Fig. 14). 

1 2 3 4 5 6 7 8 9 10 

Fig. 14 

A scale constructed in this way is called a logarithmic 
scale, and its usefulness for purposes of calculation was first 
pointed out by Edmund Gunter* in 1620. It enables us 
to find a line-segment equal in length to the logarithm of 
any number between 1 and 10. It is easy to see how, by 
means of such a scale and a pair of dividers, multiplication 
and division may be reduced to the simple graphical 
processes of adding and subtracting line-segments. 

30. The slide rule. Some years before 1630, William 
OuGHTREDf noticed that the use of the dividers might be 
avoided by constructing two equal logarithmic scales (Scales 

* Professor of astronomy in Gresham College, London (1581-1626). 
f Oughtred (1575-1660) was a fellow of King's College, Cambridge. 



THE SLIDE KULK 



U3 



A and B of Fig. 15), capable of sliding by each other, as 



indicated in the figure.* 



r> (> , 8 s id 

J 1 1 — I — i — u 



3 4 



7 8 \) 10 



Fig. 15 

The use of this simple bit of apparatus for the purpose of 
multiplication and division will be apparent from the follow- 
ing examples : 

To multiply 2 b>/ 3. Place scale B in such a way that its left-hand 
index (i.e. the division marked 1) falls directly under the division 
marked 2 on scale A. Directly above the division marked 3 on scale B, 
we shall find, on scale A, the product which (of course) is 6. To justify 
this process it suffices to note that it is equivalent to adding the log- 
arithm of 3 to that of 2. 

Fig. 15 shows scales A and B in the proper position for the purposes 
of this example. 

To divide 6 b;/ 3. Under the division 6 of scale A, place division 3 of 
scale B. Over the division 1 of scale B we shall find the quotient (f = 2) 
on scale .4 (cf. Fig. 15). 

The instrument actually in use, the Mannheim slide rule, 
is a slight amplification of the one just described (cf. 
Fig. 16). It has four scales, usually denoted by A, B, (7, D, 
respectively, the scales A and D being on the rule, and B 
and C on the slide. 




Fig. 16 



The scale A is composed of two logarithmic scales such as 
that of Fig. 14, so that its right-hand end might be labeled 
100. since log 100 = 2. On most slide rules, however, the 
first principal division on scale A after 9 is not labeled 10, 



•Obghtred'a instruments were described in publications oi William Fos- 

tf.k, one <>f hi- pupils, in 1*'> '>'-' and 1633. 



64 CALCULATION BY LOGARITHMS 

as in Fig. 14, but 1, the next one is not labeled 20, but 2, 
and so on to the last one, which is again labeled 1 instead 
of 100 or 10. Thus, the two halves of scale A are exact 
copies of one another. This is done for precisely the same 
reason that the mantissas only are printed in our tables of 
logarithms. The slide rule also makes use of the mantissas 
only. The characteristics, or what amounts to the same 
thing, the position of the decimal point in the result, must 
be obtained by inspection or by special rules. 

Scale B is on the upper edge of the slide, in direct con- 
tact with scale A on the rule, and is an exact copy of scale 
A. These two scales together may be used for multiplica- 
tion and division as explained above. 

Scale D is on the lower part of the rule. It is a single 
logarithmic scale, from 1 to 10, of the same length as the 
combined two scales of A. The logarithm of any number 
is therefore represented, on scale i>, by a distance twice as 
great as that which represents the logarithm of the same 
number on scale A. It follows from this that the number 
which is found on scale A, vertically above any number of 
scale _D, is the square of the latter. Any number on scale 
i), on the other hand, is the square root of the number ver- 
tically above it on scale A. 

Scale O is on the lower edge of the slide, in direct contact 
with slide D on the rule. It is an exact copy of scale D. 
These two scales together may be used for multiplication 
and division, according to the same rules which hold for 
scales A and B. 

Besides these four scales, the slide rule is supplied with a 
runner (cf. Fig. 16), which is useful in performing com- 
pound operations, and also in comparing two scales (such as 
A and D), which are not in direct contact with each other. 
The runner was made a permanent feature of the slide rule 
by Mannheim in 1851.* 

* Ame'de'e Mannheim (1831-1906), a distinguished geometer of recent times. 
The runner had however been used occasionally, long before Mannheim, by a 
number of English mathematicians. 



THE SLIDE RULE 65 

It often happens, in manipulating the slide rule, that the 

result is to be sought opposite a number of the slide which 
falls outside of the scale on the rule. In such cases, we may 
shift the slide, bringing the right-hand index to the place 
which the left-hand index occupied previously, and read off 
the result as before. For, such a shift has no influence on 
the mantissa, since it merely amounts to dividing the result 
by 10. On the Mannheim rule, this shifting of the slide 
may be avoided by working with scales A and B rather than 
with C and I). Scales C and i), however, have the advan- 
tage of greater accuracy. 

CO J 

If the slide be withdrawn entirely, it will be found to 
have three other scales on its reverse side, two of which are 
labeled S and T. These are scales of logarithmic sines and 
tangents, respectively, and may be used for calculating such 
products as 

c sin A, c tan A. 

The middle scale on the reverse side is used for finding 
the value of the logarithm of a number, and is important if 
we wish to compute a power of a number with a complicated 
fractional exponent. 

For more complete information concerning the slide rule, 
we must refer to the manuals which are usually presented 
to the purchaser of such an instrument.* Cheap slide 
rules, especially constructed for the beginner, may now be 
obtained of all dealers under the name Student's or Col- 
lege Slide Rule. Engineers and computers use the slide 
rule so extensively that the student will find it advis- 
able to make himself familiar with the instrument by actual 
use. 

The Mannheim slide rule, which we have described, admits 
of three-figure accuracy. In some (exceptional) cases, 
results correct to four decimal places may be obtained by its 
use. The Thacher and Fuller slide rules, more compli- 
cated instruments, but constructed on essentially the same 

* See also Raymond's Plane Surveying. 



66 CALCULATION BY LOGARITHMS 

principles, admit of far greater accuracy. The Eichhorn 
Trigonometric Slide Rule was invented for .the purpose of 
solving triangles, and is especially adapted for this work. 
But, of course, it has not the wide range of usefulness of the 
ordinary slide rule. 



CHAPTER VI 



APPLICATION OF LOGARITHMS TO THE SOLUTION OF 
RIGHT TRIANGLES 

31. The general method. We have shown, in Chapter III, 
how to solve right triangles by means of the natural functions, 
and we have become acquainted with the theory and use of 
logarithms in Chapter V. To solve a right triangle by loga- 
rithms, it suffices to combine the results of these two chapters. 
We use the same formulae as in Chapter III, but perform 
the multiplications and divisions b}' means of logarithms, 
using the table of logarithmic sines, cosines, etc., in place of 
the table of natural functions. 

In order to illustrate the various practical questions which 
arise in such a calculation, we shall give a rather extended 
discussion of the following example : 

Kxample. The legs of a right triangle were found to be 
a = 527.38 feet and b = (321.24 feet. Calculate the hypote- 
nuse and the acute angles ^L and B. 

32. The preliminary graphic 
•olution. 

We first make a drawing, approxi- 
mately to scale, making 

a = .").3 centimeters, 
b = 6.2 centimeters. 
We find by measurement 

c = 8J centimeters, 
A = In .:.. B = 49 .5. 




Fig. 17 



This figure and these measurements serve two purposes. 
In the first place, the figure helps us pick out the formulae 
which we shall need for the trigonometric solution of our 
triangle. In the second place, comparison of the approxi- 
mate values of the unknown quantities obtained graphically, 



68 SOLUTION OF RIGHT TRIANGLES BY LOGARITHMS 

with the final results as obtained by calculation, constitutes 
a valuable check. If the results obtained by the two methods 
should differ by more than can be accounted for by the in- 
accuracies of the graphic solution, we must look for a mis- 
take in our calculation. 

33. The gross errors. Mistakes, which are large enough 
to be detected by means of a graphic check, are known as 
g?*oss errors and are usually due to one of the following 
causes : 

1. The use of a wrong formula. 

2. A misplaced decimal point or, what amounts to the 
same thing, an erroneous characteristic. 

3. The use of a number taken from a wrong column in 
the tables, resulting, for instance, in erroneously using the 
log cos of an angle in place of the log sin. 

4. Addition of two logarithms when subtraction is re- 
quired or vice versa. 

5. Purely arithmetical errors of addition and subtraction. 
The errors of the first four classes can be avoided by the 

exercise of a sufficient amount of care. The student should 
not attempt to gain speed in calculation until he has first 
learned to be accurate. The errors of the last class are 
quite unavoidable, but they will usually be detected almost 
as soon as made if the ordinary arithmetical checks for 
addition and subtraction be applied every time that one of 
these operations is used. 

34. Selection of formulae and checks. After completing 
the approximate graphical solution of a triangle, we pick out 
the formulas which we wish to use in the computation. 

In our example, these are the following : 

(1) tan .4 =-, c = — — = — — , 5 = 90°-^, 

b sin A cos A 

or, in logarithmic form, 

(2) log tan A = log a - log b, B = 90° - A, 

log c = log a — log sin A = log b — log cos A. 

We have two formulae for c, and in most cases it makes little differ- 
ence which one we decide to use. If we were to use both, the agreement 



SKELETON FORM OF SOLUTION 69 

of the two results would constitute a partial check on the accuracy of our 
work. It would not be a total check however; a mistake in the Loga- 
rithm of a or b could no! he detected by means of it.* It is bardly 
worth while therefore to use both formulas for c. That one is to be pre- 
ferred which has the greater denominator, as the result obtained from it 
is likely to be the more accurate. 

For a complete check, we may make use of the equation 
o* + &a = c a , 
which, however, we prefer to write in the form 
(3) a = Vc 2 - lr = V(c- + b) (c -b), 

which is more convenient for logarithmic computation. 

It should not be necessary to write the formulas in the logarithmic 
form (2). Form (1) is shorter and more directly connected with the 
geometry of the problem. Moreover, it contains all of the information 
that is necessary for the solution of the problem, for anybody who has 
studied logarithms. 

35. The framework or skeleton form. Having selected 
the formulae, we proceed to plan the details of the computa- 
tion by providing a definite, properly marked place for every 
number which will be needed in the course of the work. 
Moreover, we shall plan these details in such a way that those 
numbers which are to be combined by addition or subtraction 
will have their places in the same vertical column next to 
each other. 

In our particular example we may adopt the following framework: 



Given 


(1) log ft = 


(4) 


I b = 


(2) log cos A = 


(7) 


\oga = 


(3) logc = 


(4) -(7) = (8) 


g // = 


(4) 


(9) 


log tan .4 = 


(3)-(4) = (5) &= 


(2) 


■'■' = 


(6) c-b= 


(9) -(2) = (10) 


Results ' /; = 


(16) c + b = 


(9) + (2) = (11) 


I c = 


(9) \og(c-b) = 


(12) 




log (c+b) = 


(13) 




log(c 2 -A 2 ) = 


(12) + (13) = (14) 




Check J log V( C « -/,*) = 


K14) = (15) 




log a — 


(3) 



Since rach a mistake could be interpreted as Leading to the correct solution 

of a triangle different from the given one, namely, that one whose Bides "' and h' 
have as logarithms the values which were, by mistake, assigned to a and h. 



70 SOLUTION OF RIGHT TRIANGLES BY LOGARITHMS 



The numbers in parenthesis merely indicate the order in which this 
skeleton form may be filled in, and how some of the results are obtained. 
The student should use these numbers only to aid him in understanding 
the construction of the framework and the plan of the computation. 
They should not be used in writing out the actual calculation. 

36. The computation. We are now prepared to carry out 
the computation. This should be done on paper ruled into 
squares of such a size that each figure may conveniently 
occupy one square. We obtain the following results : 



Given 



16 



527.38 
621.24 



log a = 2.72212 
log b =2.79326 
log tan A = 9.92886-10 
A = 40° 19'.69 
B = 49° 40'.31 
c = 814.92 



log b = 2.79326 

log cos A = 9.88215 - 10 

logc = 2.91111 

c= 814.92 

b = 621.24 

c~b= 193.68 

c + b = 1436.16 

log (c - b) = 2.28709 

log (c + b) = 3.15720 

log (c 2 - ft 2 ) = 5.44429 

2.72215 



Check. 



log Vc 2 - 6 2 

loga = 2.72212 

Remark. We observe that the check is not absolute. The agreement 
is as close, however, as we should expect. The inevitable inaccuracies, 
arising from the neglected higher decimal places, often manifest them- 
selves by discrepancies amounting to several units of the fifth decimal 
place. Consequently, we may declare the check to be satisfactoiy.* 

37. Revision of the computation when the check is un- 
satisfactory. If, in the solution of such an example, the 
results fail to check satisfactorily, the magnitude of the dis- 
crepancy will help us to locate the error. If the discrepancy 
is very great, the error must be one of the gross kind which 
we have discussed in Art. 33. In case of a comparatively 
small discrepancy, our error is probably due to one of the 
following causes : 

1. Purely arithmetical errors of addition and subtraction 
in the last few decimal places. 



* If a and b differ considerably, use as check a = V(c — b)(c + b) or 
= V(c — a)(c + «)» according as b < a or b > a. 



EXERCISES ON RIGHT TRIANGLES 



71 



2. Inexact interpolation, which would ordinarily affect 
only the last decimal place. 

3. Addition of the correction obtained by interpolation 
when it should be subtracted, or vice versa. 

This last mistake may be avoided by carefully inspecting 
the table after the interpolation has been completed, so as to 
make sure that the quantity calculated actually lies between 
the two numbers of the table between which it should fall. 



EXERCISE XV 

In eacli of the following examples (1-10), two parts of a right 
triangle are given in the usual notation. Find the other parts : 

1. c = 627. A = 23° 30'. 6. a = 13.690, b = 10.926. 

2. c = 934, £ = 76° 25'. 7. a = 67.291, c = 110.970. 

3. a = 637, A = -4° 35'. 8. b = 618.42, c = 1843.70. 

4. h = 48.532, B = 36° 44 / .00. 9. a = 965.24, .4 = 75° 15'.2. 

5. a = 38.313, b = 19.522. 10. a = 7.3298, b = 6.1032. 

An isosceles triangle may be divided into two equal rig&it triangles 
by dropping a perpendicular from the vertex to the base. Using the 
notations of Fig. 18, find the missing sides and angles 
of the following isosceles triangles. (Exs. 11-13.) 

11. h = 2.1452. B = 121° 14'.60. 

12. A = 52° 10'.2, a = 600.20. 

13. h = 7.447, A = 76° H'.OO. 

14. Prove that the area S of a right triangle is 

5 = \ be sin A — \ ac cos A . 

15. Prove that the area S of a right triangle is 

S = \ c 2 sin A cos A . 

16-25. Find the area of each of the right triangles in Exs. 1-10. 

2&-30. Find formulae for the area of the isosceles triangle of Fig. 18, 
in terms of h and h : a and h : a and h ; a and B ; a and A. 

30-32. Apply the results of Exs. 26-30 to find the areas of the isosceles 
triangles of Exs. 11-13. 

33. Show that the perimeter p of a regular polygon of n sides in- 
scribed in a circle of radius R is 

p = 2nJ28inl^, 

n 




72 SOLUTION OF RIGHT TRIANGLES BY LOGARITHMS 

that the radius r of the inscribed circle is 

r> 180° 

r = R cos •, 

n 

and that the area S of the polygon is 

e -oo • 180° 180° 

S = nR 2 sm cos ■ . 

n n 

34. Find the radius of the inscribed circle, the perimeter and the area 
of a regular pentagon, if the radius of the circumscribed circle is 12 feet. 

35. Find the perimeter, the length of one side, the radii of the in- 
scribed and circumscribed circles of a regular octagon whose area is 24 
square feet. 

36. Since the polygon of Ex. 33 approaches the circle of radius R as 

180° 180° 

limit when n grows beyond all bound, what limits do n sin cos 

n n 

180° 

and 2 n sin , respectively, approach ? 

n 

38. Applications to simple problems of surveying, navigation, 
and geography. The connection between surveying and 
trigonometry is so obvious as to require no further explana- 
tion. Moreover, we have already discussed this relation in 
Chapter 1. 

Many of the following examples are concerned with simple 
problems of surveying, and most of the technical terms which 
occur in them are self-explanatory. Nevertheless, we shall 
give a brief discussion of these terms, so as to make the appli- 
cations seem more concrete and vivid. The student who 
wishes to know more about the subject should consult a 
treatise on surveying.* 

A plumb line is a cord to one end of which is attached a 
weight. If such a plumb line be suspended, by fastening the 
other end of the cord to a fixed support, it will oscillate to 
and fro, and finally come to rest in its position of equilibrium, 
which is called the vertical line of the place of observation. 

Since the earth is approximately spherical in shape and the 
plumb line points toward the earth's center, the vertical lines 
of different places are not parallel. But the angle between 

*For instance, Raymond's Plane Surveying. 



PROBLEMS OF SURVEYING AM) NAVIGATION 



3 



the vertical lines of two stations which are not very far 
apart (say ten miles), is so small that for most purposes these 
lines may be regarded as parallel. When a tract of land (to 

be surveyed) is comparatively small, it is therefore legitimate 
to neglect the effect of the earth's curvature, and the problem 
becomes one of plane surveying. The more difficult problems 
connected with a geodetic survey, in which the earth's 
spherical form is taken into account, require knowledge of 
the methods of spherical trigonometry. 

We are concerned witli plane surveying only, so that we 
shall regard the vertical lines of all places which occur in 
such a survey as parallel. 

A vertical plane is one which contains a vertical line. 

A horizontal line or plane is one which is perpendicular to 
a vertical line. 

An inclined line or plane is one which is neither vertical 
nor horizontal. 

An angle is said to be a horizontal or vertical angle, accord- 
ing as the plane of its sides is a horizontal or vertical plane. 
Both horizontal and vertical angles may be measured by 
means of the transit (Art. 2). Inclined angles may be 
measured by means of an instrument known as a sextant. 



Horizontal line 




D=Angle of depression 

Fig. 19 



Horizontal line 
■E"=Angle of elevation 
Fig. 20 



The angle which the line of sight from the observer to an 
object makes with a horizontal line, in the same vertical 
plane, is called the angle of elevation or the angle of depres- 



74 SOLUTION OF RIGHT TRIANGLES BY LOGARITHMS 



sion, according as the object is above or below the horizontal 
plane of the observer (cf. Figs. 19 and 20). 

The angle subtended by a line is that which is obtained by 

joining the extremities of the line to the eye of the observer. 

The direction or bearing of any horizontal line is usually 

described by means of the angle which it makes with the 

north-south line or meridian, the 
latter being located approximately 
with the help of a surveyor's 
compass. Surveyors always meas- 
ure the bearing of a line as an 
acute angle from the north or south 
end of the meridian toward the 
east or west point, as the case may 
be. Thus, in Fig. 21, the bearing 
of OA is N 45° E, that of OB is S 
10° W, that of 6><7isN 30° "RT 
Surveyors usually measure dis- 
tances by means of a Gunter's chain, which is 4 rods or 66 
feet long, and is divided into 100 links. For this reason, the 
operation of measuring the length of a line in the field is 
frequently called chaining. 

In order to measure the difference of level between two 
places, A and B (cf. Fig. 22), the observer at first makes 
his telescope point in a horizontal direction by means of a 
spirit level attached to the telescope. An assistant holds a 
graduated rod, i2, in a vertical position at A, and the ob- 
server at reads, by 

Line of Sight 



means of the telescope, the 
division on the graduated 
rod where it is struck by 
the horizontal line of sight. 




Fig. 21 



Fig. 22 



He repeats this operation with the rod at B. The difference 
between the two readings gives the difference of level between 
A and B. Of course, if the difference of level between A and B 
exceeds the length of the rod, intermediate stations must be 
introduced. This operation is known technically as leveling. 



PROBLEMS OF SURVEYING A.ND NAVIGATION 75 

The navigator does not always express bearings in the 
same language as the surveyor. He divides the circum- 
ference into 32 equal parts, 
called points of the com- 
pass. Thus one point of 
the compass is an angle of 
11^ degrees. The division 
points are named, as indi- 




Fig. 23. 



The points of the Mariner's 
Compass 




cated in Fier. 23, with obvious reference to the four cardinal 
points of the compass, — north, south, east, and west. 

The navigator also makes use of the terms departure (to 
denote the east and west component of a course), and differ- 
ence in latitude (to denote the north and south component). 
These terms are illustrated in Fig. 24. 



EXERCISE XVI 

Id solving the following problems, the student should exercise his 
judgment in regard to the number of decimal places to be used in the 
calculation. (See Chap. 1. Arts. 1, 2; Chap. IV, Arts. 14, 15.) Many 
of these problems may be solved by means of three place tables (Tables 
IX. X. and XI of our collection), or by means of the slide rule. (See 
Chap. V, Arts. 29, 30.) In all of the examples the slide rule maybe 
used as a check. 

1. At a point 180.00 feet away from the base of a tower and va the 
Bame horizontal plane with it, the angle of elevation of the top was found 
to be <;.V 1 1 )'..">. Find the height of the tower. 

2. From the top of a cliff 120 feet above the level of a lake, the angle 
of depression of a boal was found to be 27° 40'. What is the air line 
distance from the top of the cliff to the boat? 



76 SOLUTION OF RIGHT TRIANGLES BY LOGARITHMS 

3. In order to measure the width of a river, a base line A C is meas- 
ured along one bank 215.6 feet long. By means of a transit, a point B 
is located on the opposite bank such that A CB is a right angle. The 
angle BAC is found to be 55° 16'.2. What is the width BC of the 
river ? 

4. From the top of a mountain 2653 feet above the floor of the 
valley, the angles of depression of two farmhouses in the level valley 
beneath, both of which were due east of the observer, were found to be 
25° and 56°. What is the horizontal distance between the two houses ? 

5. From the top of a hill, the angles of depression of two consecu- 
tive milestones on a straight level road, running due south from the ob- 
server, were found to be 22° 31' and 48° 15'. How high is the hill? 

Hint. Treat this problem as one involving two unknowns : 1st, 
the height of the hill; 2d, the horizontal distance from one of the 
milestones to the foot of the perpendicular dropped from the top of the 
hill to the horizontal plane of the road. 

6. Three lighthouses A, B, C, are situated as in 
Fig. 25, the triangle ABC being right-angled at A. At 
the moment when a ship S is crossing the line AC, the 
angle A SB is found to be 75°. If the distance between 
the lighthouses A and B is 12 miles, what are the dis- 
tances from S to A and B ? 

7. A light on a certain steamer is known to be 35 
feet above the water. An observer on the shore, whose 
instrument is 5 feet above the water, finds the angle of 

elevation of this light to be 5°. What is the distance from the observer 
to the steamer? 

8. What angle does a mountain slope make with a horizontal plane, 
if it rises 200 feet in a horizontal distance of one tenth of a mile ? 

9. The cable of a captive balloon is 835 feet long. Assuming the 
cable to be straight, how high is the balloon when all of the cable is out 
if, owing to the wind, the cable makes an angle of 25° with a vertical 

line ? £' s 

10. A ship is sailing due west at the rate of 8.9 miles 
per hour. A lighthouse is observed due south at 10 p.m. 
The bearing of the same lighthouse at 11 : 55 p.m. was 
S. 34° E. Find the distance from the lighthouse to the 
ship at the time of the second observation. 

Hint. In Fig. 26, £ and S' represent the two positions 
of the ship and L represents the lighthouse. Angle 
SS'L = 90° - 34°. Fig. 26 





APPLICATIONS INVOLVING RIGHT TRIANGLES 77 

11. Find the area of the fcracl of land corresponding to the following 
description. From A the boundary line runs X. 24 E. 20 chains to />'. 
thence X. v "> W. 35.67 chains bo C, and thence back to .1. 

12. The shadow of a chimney, 50 feet high, is 60 feet long. What is 
the altitude (or angle of elevation) of the sun at that instant ? 

13. The last row oi Beats in a circular tent is 20 feet away from the 
central pole, which is 18 feet high, and which is to be fastened by ropes 
from its top to stakes driven in the ground. How long must these ropes 
be in order that they may be 6 feet above the ground over the last row 
of seats, and at what distance from the center must the stakes be 
driven? 

14. How long must a ladder be to reach a window 45 feet high, di- 
rectly above a porch 15 feet high, if the porch projects 10 feet from the 
building? 

15. A building 125 feet high, with a flat roof, faces north on a boule- 
vard. The distance from this building to the one directly opposite is 
180 feet. How far back from the edge of the roof should a chimney 6 
feet high be placed, so as to be invisible from any point on the boule- 
vard due north of the chimney? 

16. A cylindrical pipe 36 inches in diameter is to be joined to a 
second cylindrical pipe 18 inches in diameter. The axes of the two 
cylinders are pieces of the same horizontal line and their ends are 6 feet 
apart. The joining piece is to be in the form of a frustum of a cone. 
Draw a sectional view of the joining piece and compute the length of 
the slanting side and the angles. 

17. We wish to construct a house with a gable roof. If the house is 
25 feet wide, if the height under the eaves is 27 feet and the height to 
the ridge pole 35 feet, how long must the rafters be so that their ends 
may be at a horizontal distance of 2 feet from the side of the house ? 

18. The angle of elevation of the center of a spherical balloon 20 feet 
in diameter was found to be 65 D . The angle which it subtended at the 
same time was 2° 30'. What is the height of the balloon above the 
horizontal plane of the observer? 

19. Two stations, A and B, are to be connected 
by a railroad. Both stations are in the same hori- 
zontal plane, and II is 35 miles northeast of A. The 
two stations are separated by a lake, which termi- 
nates at a point (\ 12 miles north and 20 miles east 
of .1. Find the lengths and bearings of the two ** Fig. 27 
portions of the road from A to C and from C to B. 

20. A lecture room, 50 feet long and 18 feet high, is to be supplied 
with a sloping floor. The front part of the floor, for the first ten feet, is 




78 SOLUTION OF RIGHT TRIANGLES BY LOGARITHMS 

to be horizontal so as to admit of the placing of lecture tables and ap- 
paratus. The highest part of the sloping floor, at the back of the room, 
is to be 8 feet from the ceiling. What length of sloping timbers is re- 
quired for this construction? Each of these timbers is to be supported at 
both ends and by six intermediate uprights placed at equal horizontal 
distances from each other and from the end supports. How far should 
each of these eight supports project above the horizontal part of the 
floor? 

21. In order to determine the height of a mountain above a level 
plane, we may measure a horizontal base line of length b in the same 
vertical plane with the summit of the mountain and observe the angles of 
elevation, A and B, of the summit from the two ends of the base line. 
Find a formula for the vertical height h of the mountain above the level 
of the plane. 

22. Apply the formula of Ex. 21 to the case b = 100 feet, A = 30°, 
B = 35°. 

23. A flagstaff, known to be h feet in length, stands on top of a cliff. 
An observer, in the same horizontal plane with the base of the cliff, finds 
the angles of elevation of the top and bottom of the flagstaff to be A and 
B respectively. Find a formula for the height of the cliff. 

24. Apply the formula of Ex. 23 to the case h =25, A = 40 r 25', B = 
37° 10'. 

25. The angle of elevation of the top of a spire from the third floor of 
a building was 35° 10'. The angle of elevation from a point directly 
above, on the fifth floor of the same building, was 25° 33'. What is the 
height of the spire and its horizontal distance from the place of observa- 
tion, if the distance between consecutive floors is 12 feet and the first floor 
rests on a basement 5 feet above the level of the street ? 

26. Prove the following statement. If R is the radius of the earth 
regarded as a sphere, the radius of the parallel of lati- 
tude which passes through a place P of latitude L, is 

r = it cos L. 

Hint. Use Fig. 28, where O denotes the earth's 

center, N and S the north and south poles, OE = R 

the radius of the equator, Z EOP = L the latitude of 

the place P, and r = MP the radius of the parallel 

of latitude which passes through P. 

* Fig. 28 

27. Let d be the length, in miles, of a degree of 

longitude at the equator. Show that the length of a degree of longi- 
tude, at latitude L, will be d cos L, 




APPLICATIONS INVOLVING RIGHT TRIANGLES 




28. Show that the radius r (in feet) of the horizon of an observer, h 
feet above the earth's surface, is given by the formula 

r= irb v * <•-* + *>• 

if i? denotes the earth's radius expressed in feet. 

Hint. Use Fig. 29, where OQ = OM = R. 
A£P = h t QN = r, using the angle NOQ — A as 
auxiliary. FlG ^ 

29. A micrometer screw is to be cut from a 

cylindrical steel rod. 5 millimeters in diameter, in such a way that one 
complete revolution of the screw will move the wire 
W t attached to the movable frame F (Fig. 30), 
through a distance of one millimeter. What angle 
will the thread of the screw make with a plane per- 
pendicular to the axis of the screw? 

30. A street railway track is d feet from the curbstone. In passing a 
corner (Fig. 31), where the street is deflected through an angle of K°, it 
is desired to have the rail pass at a distance A D 

of d' feet from the corner. Show that the 
radius of the circular curve A CB must be 

d — d' cos — 



H e Jane j] 



Fig. 30 



K 
cos — 

•) 



31. Solve the problem of Ex. 30 numeri- 
cally for the cases d = 10 feet, d' = 4 feet, 
A' = 90 3 ; and d --= 9 feet, d' = 3.5 feet, K = 60°. 




Fig. 31 




Fig 



32. In order to rind the horizontal distance between the points .4 and 
F which are at different levels and situated on opposite sides of a rolling 

valley (see Fig. 32), the dis- 
tances AB = d v EC - rfg, 
• •• EF = r/ 5 are measured along 
the ground by chaining. A 
transit is placed at O, and 
A'B'C .- F' represents the 
line of sight of the instrument. This line of sight is made horizontal 
by means of a spirit-level attached to the telescope. The vertical 
distances 

J.!' = /,,. BB' = h,. CO = h, DTV = h r EE 1 = h : , FF = h 6 
are measured by a rod. (Cf. Fig. 22 for method of using rod.) 
Show that the distance 

A ' F' = d x cos f, + d. 2 co> i a + d . ros ''., +d 4 cos i 4 + d :> cos i 5 , 



80 SOLUTION OF RIGHT TRIANGLES BY LOGARITHMS 

where i v i 2 , i s , i 4 , i 5 are the angles of inclination of AB, BC, CD, DE, EF, 
respectively. The angle i x is determined by the equation 



The angles i 2 ••• i 5 are determined in similar fashion. 

33. In constructing a telegraph line across a hill ABC, ••• / (Fig. 33), 
posts were set at A, B, C, ••• /, these points being determined by level 

chaining* in such a way 'that 
the horizontal distance be- 
tween any two of them A'B' = 
B'C = CD' = ... = HT = d 



feet. By leveling, the eleva- 
tions of the points A, B, C, etc., 
were found to be 




AA> 

CC 



h v BB> = h 2 , 
h 3 , ■■■ II' =\ 



Find a method for computing the amount of wire required between A 
and /, assuming that the telegraph poles are vertical and of the same 
height, and making no allowance for sag. 

39. Right triangles of unfavorable dimensions. If the 

hypotenuse and one side (say b and c) are given, we have 
the equation 

(1) cos A = - 

* e 

to determine the angle A. But, if b differs very little from 
<?, the value of cos A will be very close to unity and, as we 
observed in Art. 28, it will be impossible to determine A with 
any degree of accuracy from this equation. 

A surveyor will usually (not always) be in a position to 
avoid this difficulty. For he has a certain amount of liberty 
in the choice of his triangles. But, in many problems of 
astronomy and mathematical geography, no such choice is 
possible, so that it becomes a matter of practical importance 
to find a formula for determining the angle A, which shall 
not be liable to the same objection as (1). 



*In level chaining, the surveyor's chain or tape is held in a horizontal position, 
so as to measure the horizontal distance between two points and not" the distance 
along the slope. 




APPLICATIONS INVOLVING RKillT TRIANGLES 81 

Such a formula may be obtained as follows. In Fig. 34, 
draw AD. the bisector of the angle A and also BE perpen- 
dicular to AD. Then 

I A = A CAD = Z CBE 

(both angles being complementary to 
Z .4i7£). and / 

Fig. 34 

AE= AB = er, C#= c-5. 
Consequently we find, from the right triangle BCE, 

But 

a = Vc i2 - £ 2 = V(c - b)(c + 6), 

so that we may write, in place of (2), 



(3) tan I A = 



; c + b 



This is the desired formula, which should be applied in- 
stead of (1), whenever the value of b is very close to that of 
<•, i.e. whenever the angle A is very small. 

Some of the following examples will illustrate the useful- 
ness of this formula as well as the application of Table III 
for the functions of small angles. 

EXERCISE XVII 

1. At what distance may a mountain 14,000 feet high be seen at sea, 
if the earth's radius is 3963 miles? 

2. How high above the earth's surface must a halloon rise, in order to 
enable an observer to see a point 50 miles away ? 

3. If the moon's parallax (the angle which the earth's radius sub- 
tends a< seen from the moon) is o7 ', and if the earth's radius is 3963 
miles, what is the moon's distance from the earth ? 

4. If the angular diameter of the moon, as seen from the earth, is 
31' 20'' and the distance from the earth to the moon is 239.100 miles. 
what is the moon's diameter in miles? 

5. If the distance from the earth to the sun is 92,000,000 miles, and 
the angular diameter of the sun as seen from the earth is 32', what is 
the diameter of the sun in miles? 



CHAPTER VII 

THEORY OF OBLIQUE TRIANGLES 

40. The area of an oblique triangle in terms of two of its 
sides and the included angle. 

Let the sides 6, c of a triangle and the angle A be given. 

If we consider the side AB = c as the base, then the alti- 
tude CD=h is the length of the perpendicular dropped from 
C to AB. The foot D of this perpendicular may fall on the 

° ^f? ° 

y^Zv h 





A c D B A c B D D A c B 

Fig. 35 Fig. 36 Fig. 37 

line segment AB (Fig. 35), to the right of B (Fig. 36), or 
to the left of A (Fig. 37). In all of these cases we have, 

(1) S = i ch, 

if S denotes the area of the triangle. 

Now h can be expressed in terms of the given quantities, 
5, <?, and A, as follows. 

In Figs. 35 and 36, in which A is an acute angle, we have 

- = sin A or h = b sin J., 
b 

and therefore, by substituting this value of h in (1),- 

(2) S = J be sin A. 

If A is an obtuse angle (Fig. 37), we have 

\ = sin Z DAO = sin (180° - A), or h = b sin (180° - A\ 
b 

and consequently, by substituting this value of h in (1), 

(3) S= I rbc sin (180° -A). 

82 



AREA OF AN OBLIQUE TRIANGLE 83 

Thus, the area of a triangle is equal to 

I be sin A or .] be sin (180° - A) 

according as A is an acute or an obtuse angle. 

While we have found a complete solution of our problem, 

the result is not quite as convenient as it might be, since we 

have two different formulae for S according as A is an acute 

or an obtuse angle. Is there any way in which we might 

avoid the distinction between these two cases? 

The expression , . . . 

f be sin A 

is meaningless, from our present point of view, if A is an 
obtuse angle. For we have, as yet, given no definition for 
the sine of an obtuse angle, the definitions of Art. 7 being 
applicable to acute angles only. Clearly, however, it will 
be desirable to attach a meaning to the symbol sin A, also in 
the case when A is an obtuse angle, now that we are dealing 
with oblique triangles, some of whose angles may be obtuse. 

We may define the sine of an obtuse angle in any way we 
choose, so long as it is not iyiconsistent ivith the definitions al- 
ready agreed upon, and we naturally choose our definitions 
and notations in such a way as to reduce to a minimum the 
number of formulae and theorems which must be remembered. 

Now we can make a single formula do the work of both 
(2) and (3), by adopting the following definition for the sine 
of an obtuse angle. 

Tfie sine of an obtuse angle A is equal to the sine of the acute 
'fugle 180° — ^., which is supplementary to A; or in symbols, 

( 4 ) sin A = sin (180° - A). A being obtuse. 

As a consequence of this definition, equation (3), which 

gives the expression for the area of the triangle when A is 

obtuse, reduces to _ . _ 

& != | be sin A, 

so that formula (2) may be used whether A be acute or 
obtuse. The same formula is obviously true when A is a 



84 THEORY OF OBLIQUE TRIANGLES 

right angle ; for in that case sin A = 1, and the formula re- 
duces to S=\bc 

where c is the base and b the altitude. 

We therefore have a single formula 

S = 2 be sin A 

for the area of a triangle in terms of two sides and the included 
angle, whether the latter be acute, right, or obtuse. 

41. The law of sines. Since the triangle has three angles 
and three pairs of including sides, we may write three differ- 
ent expressions for the area of the same triangle, viz. : 

S=^bc sin A = \ ca sin B = \ab sin C. 

The equality of these three expressions is a very important 
fact, since it gives rise to the following relations between the 
sides and angles of any triangle : 

be sin A = ca sin B = ab sin 0. 

We may write these relations in a somewhat simpler form, 
by dividing all three members of the continued equation by 
the product abc We find in this way 

sin A s'mB sin O 



or 

(i) 

whence 
(2) 



a 



sin A sin B sin C 

a sin A a sin A b sin B 



b s'mB c sin Q c sin 
These formulas contain the so-called law of sines, whicn 
may be expressed in words as follows : any two sides of a 
triangle are to each other as the sines of the opposite angles. 

The first explicit statement and proof of the law of sines, known at 
the present day, is to be found in a treatise on trigonometry by the 
Persian, NasIr Addin, or Nasir Eddin (1201-1247 a.d.). Nasir Addln's 
treatise may also be regarded as the first in which trigonometry was 
treated as a separate science, independent of its applications to astron- 
omy. 



THE LAW OF COSINES $5 



EXERCISE XVIII 



1. What becomes o( fche law of sines when one of the angles (sav C) 

is a right angle V 

2. Prove the law of sines directly from Figs. 31, :>.'>, :><>, by comput- 
ing the value of h in each of the two right triangles into which .1 BC is 
divided by the altitude. 

3. Show that the law of sines may be used to solve the following 
problem: Given two angles of a triangle and one of its sides; to find 
the other sides and the remaining angle. 

4. The formula, sin (180° - .1) = sin .1, 

holds when A is an obtuse angle, as a consequence of the definition 
adopted for the sine of an obtuse angle. Show that the same equation 
is also true if A is an acute or right angle. 

42. The law of cosines. A generalization of the theo- 
rem of Pythagoras. We have learned to recognize the im- 
portance of the theorem of Pythagoras in the theory of 
right triangles, and the question naturally arises : what 
takes the place of this theorem in the case of an oblique 
triangle ? 

Most students will remember that the answer to this ques- 
tion is contained in the following two propositions of geom- 
etry : 

Theorem 1. In any triangle the square of the side oppo- 
site an acute angle is equal to the sum of the squares of 
the other two sides diminished by twice the product of one 
of those sides and the projection of the other upon that 
side. 

Theorem 2. In any obtuse triangle, the square of the 
side opposite the obtuse angle is equal to the sum of the 
squares of the other two sides increased by twice the product 
of one of those sides and the projection of the other upon 
that side. 

The proof of Theorem 1 (repeated from Geometry) is as 
follows : Let A be an acute angle. The triangle ABC 
will have the form represented in Figs. 38 or 3 ( .), accord- 



86 



THEORY OF OBLIQUE TRIANGLES 



ing as the angle B is acute or obtuse. In either case we 

P ut BQ= a, OA = b, AB = e, ' 

and AD = m, 

tC so that m is the projection of b 
upon c, or upon c produced. 
The right triangle BCD gives 




^ r/i j)c-mB 
Fig. 38 




t 2 = p + i?# 



in both cases. Now i?Z) = c — m in Fig. 38, and BD — m — c 
in Fig. 39. Therefore we find, in either case, 

a 2 = h 2 + c 2 — 2cm + m 2 . 

The right triangle AOD gives 

h 2 = b 2 — . m 2 

in both cases. If this value of A 2 be substituted in the 
equation above, we find 

(1) a 2 = b 2 4- c 2 — 2 em (A ^ern^r aw acw£e angle), 

which proves Theorem 1. 

To prove Theorem 2, we refer to Fig. 40. We have, in 
this case, 

a 2 = h 2 + ~B& = h 2 + (c + m) 2 

a 2 = A 2 -h c 2 + 2 cm -f m 2 , 



or 



and 
whence 

(2) 



h 2 = b 2 



m ■ 



D in J, v 

Fig. 40 



b 2 + c 2 -\- 2 cm (A 



an obtuse angle), 



which proves Theorem 2. 

From either Fig. 38 or 39 we obtain, by observing the 
right triangle A OB, 

m = AD — b cos A. 

In Fig. 40 we have instead, 

<L CAD = 180° - A, m = b cos CAD = b cos (180° - A). 

If we substitute these values in (1) and (2), we find 

(3) a 2 =P + c 2 -2 be cos A (if A is acute), . 

(4) a 2 = b 2 + c 2 + 2 be cos (180° - A) (if A is obtuse). 



THE LAWS OF COSINES 87 

Just as in Art. 41, we have found two different theorems 
and two different formulae for the two cases when A is an 
acute or an obtuse angle. Can we again find a single theorem 
and a single formula to do the work of both ? 

Equations (3) and (4) show that this may indeed be done, 
provided that we define the cosine of an obtuse angle to be a 
negative number, numerically equal to the cosine of its supple- 
ment (which is of course an acute angle). For, with this 
definition, we shall have 

(5) cos A = — cos (180° — A) (if A. is an obtuse angle), 

so that (4), as a consequence of (5), assumes the same form 
as (3). But formula (3) holds also when A is a right angle, 
for in that case cos A = 0, and the formula reduces to the 
theorem of Pythagoras. Thus, one and the same formula (3) 

a 2 = b 2 + c- - 2 be cos A 

ivill be applicable to all cases if it be understood, in accordance 
with our definitions, that cos A is positive, zero, or negative 
according as the angle A is acute, right, or obtuse. 

Equation (3) is generally known as the law of cosines, and 
completely replaces Theorems 1 and 2 of this Article. The 
law of cosines obviously enables us to compute the third side 
of an oblique triangle when two sides and the included 
angle are given. But it also enables us to find the angles of 
a triangle when its three sides are given. For, we find from 
(3), by transposition, 

2 be cos A = b 2 + c 2 — a 2 , 
and therefore 

b 2 + c 2 - a 2 



(6) cos A 



2 be 



The two geometric theorems (Theorems 1 and 2 of this article) to 
which the law of cosines is equivalent, were well known to the Ancients; 
and the problem of finding the angles of a triangle, when its three sides 
are given, was solved by Ptolemy (2d century a.d.) of Alexandria in 
his Almagest by means of these theorems. The explicit formulation of 
the law o; however, seems to be due to the great French mathe- 

matician, Francois Viete (also known as Vieta), (1540-1603). 



88 THEORY OF OBLIQUE TRIANGLES 

EXERCISE XIX 

Solve the following triangles, using the tables of squares and natural 
functions : 

1. a = 2, 6 = 3, C = 30°. 3. c = 2.34, a = 4.31, B = 116°. 

2. b = 3.5, c = 2.4, A = 52°. 4. a = 3, 6 = 6, c = 8. 

5. a = 1.0, b = 2.0, c = 1.5. 

6. The relation cos A = — cos (180° —A) is true for all obtuse an- 
gles A as a consequence of the definition of the cosine of an obtuse angle. 
Prove that this formula is also true if A is any acute angle or a right 
angle. 

7. Show that the relation sin 2 A + cos 2 .4=1 holds for obtuse as well 
as for acute angles. 

8. If A is an acute angle, 

, sin A , A cos A A 1 .1 

tan A = , cot A = — , sec A = , esc A = 



cos A sin A cos A sin A 

Let us define tan A, cot A, sec A, esc A by means of these same equa- 
tions when A is an obtuse angle. Show that, as a consequence of these 
definitions, we have 

tan ,4 = - tan (180° - A), cot A = - cot (180° - A), 
sec A = - sec (180° - A), esc A = esc (180° - A), 
for any obtuse angle A. 

9. Show that the equations of Ex. 8 are valid also if the angle A is 
acute. 

10. The law of cosines gives three equations for any triangle. 
Equation (3) of Art. 42 is one of these. Write the other two. 

11. Write the equations for cos .4, cos B, and cos C in terms of the 
three sides of the triangle. 

12. Show that in any triangle, 

a 2 + b 2 + c 2 - 2 be cos A - 2 ca cosB - 2 ab cos C = 0. 

43. Properties of the functions of an obtuse angle.* We 

have defined in Art. 40 the sine, in Art. 42, the cosine, and 
in Ex. 8, Exercise XIX, the remaining functions of an ob- 
tuse angle. As a consequence of these definitions we have 
the following system of equations : 

*Some instructors may prefer to change the order of topics by passing to the 
discussion of the general angle, Art. 60 et seq., and returning later to Art. 44. 
There is no reason why this should not he done. 



FUNCTIONS OF AN OBTUSE ANGLE 89 

rsin (180° -A) = sin A, cos (180° - A) = - cos A, 

(1) I tan (180° - A) = - tad, cot (180° - J.) = -cot A 

[sec ^180°-^) = - sec A esc (180° - 4) = csc.l. 

which arc valid whether the angle A be acute, right, or ob- 
tuse. (^Cf. Exercise XVIII, Ex. 4, Exercise XIX, Exs. 6 
and 8.) 

But an obtuse angle B may be written either in the form 

B = 180° - A, or B = 90° + A', 

both A and A' beinsr acute angles. Moreover, from 

B = 180° - A = 90° + A' 
follows 

A' = 90° -4, or A = 90°- A', 

that is, the angles A and A' are complementary. Let us 
substitute A = 90° - ^.' in (1). We find that the left mem- 
ber of the first equation of (1) becomes 

sin (180° -A) = sin [180° - (90° - A')] = sin (90° + A'). 

The right member of the same equation assumes the form 

sin A = sin (90° - A') = cos 4'. (Art. 10.) 

Consequently, this first equation of system (1) becomes 

sin (90° -f A') =cosA'. 

In the same way we may prove the remaining equations of 
the following system: 

| sin (90° + A!) = cos A', cos(90° + A') = -sin J/, 

(2) I tan (90° + .4') = -cot A', cot (90° +A')= -tan A', 
lsec(90° + 4') = -cscA', csc(90° + A')=secJ.'. 

The student should carry out the details of these substi- 
tutions, and note the close resemblance between these for- 
muke and the formulae of Art. 10, for sin (90° — A}, 
cos (90° -A), etc. 

This resemblance is so close as to make it easy to remember equations 
(2). Their right members differ from the right members of the corre- 
sponding equations for sin (90 — .1). cos (1)0 - .1), etc.. at most in 
sign. This remark suffices to help us remember that sin (90° + A) 



90 THEORY OF OBLIQUE TRIANGLES 

is either equal to + cos A or to — cos A, that cos (90° 4- A) is either 
equal to + sin A or — sin A, etc. In order to choose between these 
alternatives we may argue as follows. Let A be an acute angle. Then 
sin A and cos A are both positive. But since 90° + A will then be ob- 
tuse, sin (90° + A) is positive and cos (90 + A) is negative. Therefore, 
of the two alternatives 

sin (90° + A) = cos A, or sin (90° + A) = - cos A, 

we must discard the latter, since its left member is positive and its right 
member is negative. Similarly, of the two alternatives 

cos (90° + A) = sin A, or cos (90 + A) = — sin A, 

we must discard the former. For it involves the contradiction that the 
negative number cos (90° + A) should be equal to the positive number 
sin A. This argument fixes the two formulae 

sin (90° + A) = cos A, cos (90° + A) = - sin A 

in our memory. The remaining formulae of system (2) may be remem- 
bered in similar fashion. , 

The first two formulae of system (1), namely, 

sin (180° - A) = sin A, cos (180° - A) = - cos A, 

are easy to remember, since it was upon these equations that we based 
our definitions of the sine and cosine of an obtuse angle. The remain- 
ing formulae of system (1) follow directly from these two. 

EXERCISE XX 

1. Express the following quantities as functions of acute angles; 
sin 100°, cos 115°, tan 162°, cot 99°, sec 120°, esc 175°. 

2. By means of the table of natural functions, find sin 98°.5, 
cos 176°.3, tan 124°.7, cot 134°.6. 

3. Explain how equations (1) and (2) of Art. 43 provide two differ- 
ent methods of finding the functions of obtuse angles from the table for 
acute angles. 

44. Other formulae for the area of an oblique triangle. The 

methods of Art. 40 suffice to determine the area of a tri- 
angle if one side and the corresponding altitude (c and A), 
or if two sides and the included angle (6, <?, and A) are 
given. 

Let us suppose now that one side and two adjacent angles 



ADDITIONAL AREA FORMULA Hi 

(<?, A, B) are given. We have in the first plaee from 

Art. 40 

(1) S=ibc sin A, 

and it only remains to modify this formula by expressing b 
in terms of c, A, and i?. The law of sines, in the form 

c sm C T 
and the equation 
(3) A + B + O = 180° 

enable us to do this. For we find from (2) and (3) 

c sin B c sin B c sin B 



b = 



sin C sin [180° - (A + 5)] sin (A + £) 

since, according to Art. 43, the sine of any acute or obtuse 
angle is equal to the sine of its supplement. 

If we substitute this value of b in (1), we find 

xix p c 2 sin A. sin 2? 

(4) S = 2 sin (A + £)' 

the desired formula for S in terms of c, A, and B. 

If one side <? and two angles, not both adjacent to <?, are 
given, we may first find the third angle from (3) and then 
use formula (4). 

Let us, finally, obtain a formula for S in terms of the 
three sides a, 6, c. We start again from equation (1) which 
already contains the sides b and c of the triangle, and which 
will give the desired formula if we can express sin A in 
terms of a, b, and c. This may be done by making use of 
the law of cosines (which gives cos A in terms of a, 6, and 
'■ ». and the relation 

I 5 i sin 2 A + cos 2 A = 1, 

which holds for obtuse as well as for acute angles. (See 
Ex. 7. Exercise XIX.) 

To avoid the inconvenience of writing cumbersome square 



92 THEORY OF OBLIQUE TRIANGLES 

root signs, let us square both members of (1). We find, 
making use of (5), 

S 2 = i b 2 c 2 sin 2 A = \ b 2 c 2 (1 - cos 2 A) 
or, factoring the binomial on the right, 

#2=1 b 2 c 2 (1 + cos A) (1 - cos A). 
By the law of cosines (Equation (6), Art. 42), this becomes 

= 52 £ 2 2 5e + b 2 + c 2 - a 2 2be-b 2 -c 2 + a 2 
4 2 6c 2fo 

= iV [(6 + 2 -« 2 J [a 2 -(6-,) 2 ]. 

Each of the factors on the right member may be factored, 
giving the equation 

(6) aS' 2 = t V (b + c + a) (b + c-a) (a+b-c) ( a -b + c). 

The first factor on the right member of (6) is the perimeter 
of the triangle, and the formula becomes especially simple 
if we denote half of this perimeter by s, so that 

(7) a + b + c=2s. 

The other three factors of the right member of (6) will then 
become 

-a + b + c=2s-2a = 2(s-a), 

(8) a-b + c=2s-2b = 2 (s - 6), 
a + b-e=2s-2c=2(s-c~), 

so that (6) reduces to 

S 2 = s (s — a) (s — b) (s — e), 

whence finally, since S is positive, 

(9) S = Vs (s-a) (s-b) (s-c), 

a famous equation generally known as Hero's formula, after 
Hero of Alexandria, who lived about 120 B.C., and wrote 
a famous textbook on surveying. 



RADIUS OF DESCRIBED CIRCLE 



93 



EXERCISE XXI 
Find the area of the following- triangles to four significant figures. 

1. a = 2, h = 3, C = 30°. 4. a = 3, .1 = 30°, 5 = 75°. 

2. J = 3.5, c = 2.4, .1 = 52°. 5. a = 3, 6 = 6, c = 8. 

3. c = 5, .4 = 30°, 5 = 75°. 6. a = 1.0, 6 = 2.0, c = 1.5. 




45. The radius and center of the inscribed circle. The 
fundamental basis for all of the different formulae which Ave 
have obtained for the area of a triangle, so far, was the 
equation #=1^, 

from which all of the others were derived. 

But this formula is unsymmetrical, since it singles out one 
of the sides of the triangle and subjects it to a treatment 
different from that accorded to the 
other two. We may avoid this lack 
of symmetry by picking out a point 
M anywhere inside of the triangle 
ABC (see Fig. 41), and joining M 
to the three vertices. The area of 
the given triangle will then appear 
as the sum of the areas of the three 

triangles BMC CMA, AMB, whose altitudes are respec- 
tively equal to MB, ME, and MF. 

Clearly, there is one position of the point M which is better 
adapted for this purpose than any other, namely the center 
of the inscribed circle. For the distances from to the 
three sides of the triangle are equal to each other, so that 
the three triangles BOC, CO A, and AOB will have equal 

altitudes. 

Let (Fig. 42) be the center of 
the inscribed circle and let r be the 
radius of this circle. 

Then the area of the triangle is 
S=BOC+ CO A +A0B 



Fig. 41 




_ 1 



FlO. 4-' 



^ ar + \ hr + \ cr 
i (a + b + c) r, 



94 THEORY OF OBLIQUE TRIANGLES 

or finally 

(1) S = sr, 

where s denotes the half -perimeter of the triangle as in 
Art. 44. 

Since we also have 



# = Vs(s-a)(s-6)(s-tf) (Equation (9), Art. 44), 

we find from (1), substituting this value of 8 and dividing 
both members of the resulting equation by s, 

(2) r= J (s-a)(8-b)(s-c) } 

which enables us to compute the radius of the inscribed circle 
when the sides of the triangle are given. 

If we wish to know, not merely the radius of the inscribed 
circle, but also the position of its center, we must compute 
the lengths of the line segments AF, BF, etc. in Fig. 42. 
Now we know from Geometry that 

(3) • AF = AF, BB = BF, OF = OB. 

Moreover, the sum of these six line-segments is equal to the 
perimeter 2 s of the triangle. Therefore the sum of three of 
these segments, one chosen from each of the three equations 
(3), will be equal to half of the perimeter. That is, 

AF+ BB + OB = s. 
But 

BB + OB = a, 
and therefore 

(4) AF= AE= s- a. 

In the same way we find 

(5) (BB = BF=s- b, 
I OF = OB = s-c. 

The three equations, (4) and (5), enable us to locate the 
points D, F, F in which the inscribed circle touches the sides 
of the triangle and consequently determine completely the 
position of the center of the circle. 



RADIUS OF CIRCUMSCRIBED CIRCLE 95 

46. The half-angle formulae. Since the center of the 
inscribed circle is the point of intersection of the three 
angle bisectors of the triangle (see Fig. 42), the angle 
FAO is equal to \ A. In the right triangle FAO, we have 
therefore 

But FO is the radius r of the inscribed circle, and AF we 
have just found to be equal to s — a. (Equation (4), Art. 
45.) We find therefore 



(1) 










tan 


i'-; 


r 








and 


in 


the 


same 


way 














(2) 








tan 


l B - 


r 


tan I 


C- 


8 


r 
-c' 



the value of r being given by equation (2) of Art. 45. 

These three .equations, usually known as the half -angle 
formulae, are very important in providing a second method 
for computing the angles of a triangle when its sides are 
given. They are far more convenient for this purpose than 
the law of cosines, if logarithms are to be used. In fact, the 
law of cosines is so cumbersome from the point of view of 
logarithmic calculation, that we shall seek to find a substi- 
tute for it also in the only other casein which we have pro- 
posed to make any use of it, namely in the solution of a 
triangle when two sides and the included angle are given. 
(See Art. 42, p. 96.) 

We should not neglect to note, however, that the law of cosines has 
in recent times again come into practical use, especially in engineering 
practice, the calculations being performed not with logarithms, but with 
the help of tables of squares and products or with a calculating machine. 
The Eichhorn Trigonometric slide rule, mentioned on page 66, is based 
entirely upon the law of cosines. 

47. The circumscribed circle. The center of the circum- 
scribed circle is the point of intersection of the three per- 
pendicular bisectors of the sides of the triangle. Therefore, 



96 THEORY OF OBLIQUE TRIANGLES 

if iVis the middle point of the side AB of the triangle (see 
Fig. 43), the line ON will be perpendicular to AB. 

Now angle AOB is measured by the arc AB, and the 
angle AOB is measured by half 
this arc (being an inscribed angle). 
Therefore 

ZAON=Z AOB= O 
since each of these angles is equal 
to half of Z AOB. Consequently 
we find, making use of the right 
triangle A ON, 

fig. 43 sin C = sin A 0N= — — - . 

AO 

If we denote AO, the radius of the circumscribed circle, by 
B, this becomes 

sin<7=i^=-V ,or2i£ = 




B 2B sin C 

whence, making use of the law of sines (Equation (1), Art. 

41), 

(Y) 2 B = a = = ° 

sin A sin B sin 0' 

thus completing the law of sines by providing a simple geo- 
metrical interpretation for the common value of the three 
equal ratios a b c 

sin .A' sin J9' sin (7' 
namely, the diameter of the circumscribed circle. 
Since the area of the triangle is 

S = J be sin A, 
and since we find, from (1), 

sin A = — — -, 
2B 

we obtain the following remarkable formula 

X) ±B 

for the area of the triangle. 



THE FORM RATIOS OF A TRIANGLE 97 

EXERCISE XXII 

1. Compute the radius of the [ascribed circle and the lengths of the 

line-segments into which this circle divides the sides of the triangle 
whose sides are a = 3, b = 6, c = 8. 

2. What relation will there be between the sides of a triangle if the 
inscribed circle bisects one of its sides? If it touches one of the sides at 

a trisection point ? 

3. Find formulae for the distances from the center of the inscribed 
circle to the three vertices of the triangle. 

4. Making use of the results of Ex. 3, show that 



sin I A = \ (s — b)(s — c)/hc and cos \ A = vs(s — a) /he. 

5. Find a formula for the area of the inscribed circle, and for the 
ratio of this area to that of the triangle itself. 

6. By means of the formula? of Ex. 5, show that 

7. Show that 

r = s tan £ .4 tan \ B tan \ C. 

8. A circle is said to be escribed to a triangle, or inscribed externally, 
if it touches one of the sides externally and the prolongations of the other 
two sides (Fig. 44-). There are three such 

circles for every triangle. Let r a be the 
radius of that one which touches the side a 
externally. Show that 

.4.1/= AX = s 
and 

5 rs 



.v 




r a — s tan h A = 

s — a s — a 

9. If r a , r h . r a are the radii of the three escribed circles, and r de- 
notes the radius of the inscribed circle, show that 

r r a r b r c 
10. Show that 

'V, +r b + r e — r = ±R, 

if R is the radius of the circumscribed circle. 

48. The form ratios of a triangle. If two sides of a 
triangle are equal, the triangle is said to be isosceles ; if no 
two sides are equal, it is usually called a scalene triangle. 
But if the difference between two sides of a triangle is small, 



98 THEORY OF OBLIQUE TRIANGLES 

as compared with the combined length of these two sides, the 
triangle will differ but little from the isosceles form. Con- 
sequently, the ratio of the difference between two sides of a 
triangle to their combined length may be taken as a nu- 
merical measure for the departure of the triangle from the 
isosceles form, or as its form ratio with respect to those two 
sides. 

There are three such form ratios for every triangle. If 
we assume that the notation be so chosen that 

a>b^>c, 

these three form ratios are 

a — b a — c b — c 

a+ b a+ c b + c 

Clearly, one of them will be equal to zero if the triangle is 
isosceles, and all three will vanish for an equilateral tri- 
angle. 

By means of the law of sines, each of these form ratios 
may be expressed in terms of the angles of the triangle. In 
fact, according to Art. 47, equation (1), we have 

a = 2Bs'mA, b=2Rs'mB, c=2RsmC, 

so that we obtain the expression 

..<. a — b _ sin A — sin B 

a + b sin A + sin B 

for the form ratio with respect to the sides a and b. Simi- 
lar expressions hold, of course, for the other form ratios ; 

<^Ll and \=^. 

a+ c b + c 

49. The formulae for the sum and difference of two sines. 

The fraction which occurs in the right-hand member of 
equation (1), Art. 48, is not adapted to logarithmic calcula- 
tion. But we shall show that both numerator and denom- 
inator of this fraction, namely, the expressions sin A— sinB, 
and sin A + sin 5, may be written in the form of products, 
thus making it easy to compute their values by logarithms. 



THE SUM AND DIFFERENCE OF TWO SINES 



yy 




To discover the product form of these expressions, we 
must construct a figure in which the angles A and B, the 
sines of these angles, and the sum 
and difference of their sines, shall 
appear. 

We construct first (Fig. 45) the 
angles 

Z XOP = A and Z XOQ = B. 

With their common vertex as 
center, and any convenient radius 
r, we draw a circle whose intersec- 
tions with the sides of the angles we denote b)^ X, P, and Q, 
respectively. 

Let PM and QX be perpendicular, and QR parallel to 
OX. Then 

(1) r sin A = MP, and r sin B = XQ. 

But MP and XQ are the two bases of the trapezoid MPQX. 
If S is the middle point of the chord PQ, and US be drawn 
perpendicular to OX, US is the median of this trapezoid, and 
we shall have, 

US = I {MP + NQ\ 

whence, according to (1), 

(2) r (sui A + sinB) = MP + XQ = 2 US. 
On the other hand. 



TS= I PR 

and hence, 

(3) r(sinA-amB') 



UMP-XQ), 
MP - XQ = 2 TS. 



In order to express the right members of (2) and (3) in 
terms of r, A and B. we consider the right triangles OSU 
and SQT in which US and TS occur. It is apparent from 
the figure that 

\ASOQ= \ QO/>= I (A- B), 

U UOS=B+ I (A- R) = \ (A +£), 



(4) 



100 THEORY OF OBLIQUE TRIANGLES 

and 

(5) Z TSQ = Z UOS = J (A + £), 

since each of these angles is complementary to the angle 

uso. 

Hence we have from the right triangles 0#Z7and SQT, 

US = OS sin UOS = OS sin ±(A + B\ 
TS = SQ cos TSQ = SQ cos \(A + £), 

and from the triangle OQS, 

OS = OQ cos SOQ = r cos -| (J. - 5), 
^ > SQ= OQ sin #6>£ = r sin. J (J. - #). 

Substituting (6) and (7) in (2) and (3) and dividing by 
r, we find 

sin A + sin S = 2 sin | (^ + 15) cos J (^ -B), 

(8) i j 

sin^L - sin# = 2cos| (^ + B) sin^ (^4 - B)» 

which are the desired formulas. 

The student should observe that our proof of equations 
(8), based on Fig. 45, remains valid even if A is an obtuse 
angle, as long as MP is greater than or equal to NQ ; that is, 
as long as A + B does not exceed 180°. We may therefore 
apply equations (8) whenever A and B are two angles of the 
same triangle, since the sum of two angles of a triangle can 
never exceed 180°. 

50. The law of tangents. Let us now return to the expres- 
sion (1) of Art. 48 for the form ratio of the triangle with 
respect to the sides a and b, of which sides we shall assume 
a to be the greater. We had found 

a — b _ sin A — sin B 
a + b sin A + sin B 

If now we make use of equations (8) of Art 49, we find 

a - b = 2 cos j ( A + B) sin % (A - B) 
a + b 2 sin \ (A + B) cos l (A - B) ' 



THE LAW OF TANGENTS 101 

whence 

2^4 = cot J (A + B) tan UA - B), 
a + b 

since we have, for any angle M, 

sin M , , r cos 3i" , 7,^ 

■ = tan M, — = cot M. 

cos M sin M 

But the tangent and cotangent of the same angle are re- 
ciprocals, so that we may write finally 

a-b taul(A-B) 

(1) = i 9 

a+b tan | (A + B) 

a very important formula, generally known as the law of 
tangents. We find in the same way : 



(2) 



c _ tan I (A - (7) 



a + c tan J (^4 + (7)' 
ft- g = tan|(.g- G 7 ) 
6 + c tan K-S+ #) 



The law of tangents seems to have been expressed in this form for the 
first time by Vieta. to whom we also owe the modern form of the law of 
cosines. It had however been stated, in a more complicated but equiva- 
lent form, about ten years earlier by the Dutch mathematician Thomas 
Fink or Fixcmus (1561-1656) in his Geometvia rotundi. It was also 
Fink who first introduced the names tangent and secant for the functions 
which we now call by these names. Many previous authors had used the 
name umbra = shadow for the function which we now call tangent, on 
account of the relation of this function to the shadow cast by a vertical 
stick. (Compare Art. 3 and solve the problem there attributed to Thales 
by trigonometry.) 

Fink not only discovered the law of tangents, but pointed out its 
principal application ; namely, to aid in solving a triangle when two 
sides and the included angle are given. The possibility of such an ap- 
plication will appear from the following 

Illustrative Example. Given a. b, and C. To find A, E, and C. 
Solution. The law of tangents (Equation (1)) giv<-s 

(3) tan \{A - B)= ^\ tan \{A + B), 

a + U 



102 THEORY OF OBLIQUE TRIANGLES 

an equation in which the right member is completely known, since «, b, 
and C are given, and since 

(4) \{A +B)= 1(180° - C)= 90° — \ C. 

Thus we can compute tan l( A — B) by means of (3) and then find 
l(A — B) from the table. Hence, knowing \{A — B) and \(A + B) we 
can find 

(5) A =\(A + B)+i(A -B), 
B = ±(A +B)-\(A - B). 

The sine law now enables us to find c, since 

sa\ c sin C •„ „„ „ a sin C 

(o) - = gives c = 

a sin A sin A 

If we wish to solve the same problem by the law of cosines, we first 
compute c from the equation 

(7) c 2 = « 2 + b 2 - 2ab cos C 

and afterward determine the angles A and B by using the law of sines. 

The first method has the advantage over the second that formulae 
(3), (4), (5), (6) are in a convenient form for logarithmic computation, 
while equation (7) is not. 

51. A second proof of the law of tangents and Mollweide's 
equations. The law of tangents may also be proved directly 

from a figure 
without making 
use of the for- 
mulae for 

sin A + sin B 
and 
sin A — sin B. 

Let BO = a 
and CA = b 
(Fig. 46) be 
FlG - i6 two sides of the 

triangle ABG, and let a>b. Draw a circumference with C 
as center and radius equal to 5, and let D and E be the 
points in which this circumference meets BO and repro- 
duced. Let F be the point of intersection of the circum- 
ference with AB. Then we have 




SECOND PROOF OF LAW OF TANGENTS 103 

(1) BD = a - 6, BE = a + b. 

Since Z EC A is an exterior angle of the triangle ABC, the 
opposite interior angles being A and B, we have 

Z J0OA = A + B, 

so that 

(2) Zj0ZU = £(.A + .B) 

since the latter angle, being inscribed in the circumference, 
is measured by half the arc which it subtends. 

It remains to find an angle in Fig. 46, equal to I (A — By 
In order to do this, let us draw CF. Since A CF is an isos- 
celes triangle, we have 

Z.AFC= ZFAC=A 

and, since Z AFC is an exterior angle of the triangle BFC, 

A= Z FCB + B, 
whence 

Z FCB = A- B. 

Since Z DAF is an inscribed angle subtending the same arc, 

(3) Z DAB = Z BAF = \(A - B). 

Let us apply the law of sines to the two triangles ABB 
and ABE. We find, from the first triangle, 

a-b sin DA B sin j (A-B) = sin % (A - B) 
c sin ADB ~~ sin (180° - EDA) sin EDA 

or finally 

a-h = sin j- (A - ff) 
c sin I (A + B)' 

Similarly the triangle ABE gives 

a^^sinEAB 
c ~ sin AEB' 
But 

Z .£A£ = Z ,£AZ) + Z DA£ = 90° + Hi- 5), 
Z AEB = 90° - Z £7)A = 90° - J (A + B ). 

owing to (2) and (3) and the further fact that Z EAD is a 
right angle, since it is inscribed in a semi-circumference. 



(4) 



104 THEORY OF OBLIQUE TRIANGLES 

Therefore 

a + b _ sin [90° + 1JJJ- By] 
c sin [90° -l (A + ^)]' 

which may be written 

f^\ a + b __ cos \ {A — i?) 

c ; T""co S i(i + 5y 

as a consequence of equations (2), Art. 43, and of Art. 10. 

If now we divide equations (4) and (5), member by mem- 
ber, we find 

a - b sin j- ( A - B) cos \ (A + B) tan j- (A - B) . 
a + b sin 1 (J. + B) cos J ( J. - B) tan J (4 + B) ' 

that is, the law of tangents. 

Incidentally we have found two new formulas, (4) and 
(5). They assume a somewhat more serviceable form by 
means of the relation A + B + 0= 180°, which gives 

and therefore 

sin I (A + B) = cos 1 G 7 , cos J (A + B) = sin | (7. 

If we introduce these values in equations (4) and (5), 
they become 

a - h sm\(A-B) a + b cosf(4-JJ) 



(6) 



cosfc c sin**? 



These formulae, known as the Mollweide equations, are 
particularly convenient for the purpose of checking the accuracy 
of the numerical solution of a triangle. For each of these 
equations contains all of the six parts of the triangle, so that 
an error in any one of these parts would be likely to make 
itself felt by a lack of agreement between the two members 
of one of these equations. 



THIRD PROOF OF LAW OF TANGENTS 105 



Of course there are two other pairs of equations of the 

;6), t 

<• b + c 



form (6), the left members of which are — - — , "; ° * all( l 

b b 



respectively. 



It is not justifiable historically to call equations (6) Mollweide's 

equations. The formula for a is to be found in Newton's Aritli- 

c 

metica Universalis. Botli equations (G) are given in Simpson's Trigo- 
nometry, Plane and Spherical (1748), and also in F. W. vox Oppei/s 
Analysis Triangulorum (174(5). All of these works antedate considerably 
the publication of these equations by MoLLWEIDE in 1808. 

EXERCISE XXIII 

1. Show that the law of tangents may also be obtained from 
Fig. 40 by drawing through D a line parallel to AE and meeting the 
side AB in G, and then computing tan i (A — B) and tan \ (A + B) 
from the right triangles ADG and AED. 

2. Still another proof of the law of tangents proceeds as follows. 
In Fig. 47, draw CD. the bisector of angle C, and drop perpendiculars, 
AL and BM. from .4 and B to CD. 
Let N be the point of intersection 
of All with CD. Then 

/.BCD = ZDCA =\C 
and 

(1) /LAN=ZNBM 

= DO - ZllXM. FlG - 47 

But Z BNM is an exterior angle of the triangle BCN. so that 

Z BNM = B + \ C. 

Moreover, 

90° = l(A + B + C) 

and therefore, by substitution in (1), 

ZLAX = \(A + B+ (')- (5 + JC) =\(A - B), 

Now. from the right triangles A XL and BNM, we see that 

tan>(.l -1*)=M=^, 
} LA MB' 
and therefore also 

LN + MN 




(2) tan J (-4 - ^ 



LA + Ml: 



106 THEORY OF OBLIQUE TRIANGLES 

But 

LN + MN = CM - CL = a cos \ C - b cos 1 C = (a - b)cos | C, 

LA + MB = b sin \ C + a sin \ C = (a + b) sin \ C, 

which gives, on substitution in (2), 

tan \{A -B) = -^-|cot \ C, 
a +o 

and this is equivalent to the law of tangents, since 
C = 180°-(A +B). 



CHAPTER VIII 

SOLUTION OF OBLIQUE TRIANGLES 

52- The fundamental problem. Each of the laws found in 
Chapter VII contains four of the six parts of a triangle, and 
thus suggests the possibility of computing any one of these 
four parts when the other three parts which occur in that 
law are given. 

On the other hand, the relation 
(1) A + B + C = 180°, 

which is true for every triangle, contains only three of its 
parts. The three angles of a triangle are therefore not inde- 
pendent of each other, as is the case with the three sides. 
The familiar fact, that a triangle is not determined by its 
three angles, is to be regarded as a consequence of this. 
For. on account of relation (1), the three angles of a 
triangle are not independent data. 

Now there exists no other equation ivhich, like (1), contains no 
more than three parts of the triangle and which is true for all 
triangles. For, if there were such an equation, for instance, 
between a, b, and <?, it would be impossible to find a triangle 
in which more than two of these quantities have arbitrarily 
assigned values since two of the three quantities would then 
determine the third. But this is contrary to well-known 
facts of geometry ; since a triangle may be constructed for 
which all three sides a. b, and c have arbitrarily assigned 
values, provided only that a 4- b > c. The values of a and b 
do not determine the value of c; while the values of A and B 
do determine the value of C. 

Our illustration shows incidentally that there may, and, in fact, do 
exist, besides the equation (1), certain inequalities between three parte of 
a triangle which must be satisfied by all triangles. The inequality 
a + h •> r is an instance. 

107 



108 SOLUTION OF OBLIQUE TRIANGLES 

Any three parts of a triangle, unless all three of them are 
angles, may therefore be regarded as independent data. 
Consequently there arises the following fundamental problem: 

To find the remaining parts of a triangle when any three in- 
dependent parts are given. 

The discussion of this problem leads to a division into four 
cases. 

Case I. One side and two angles are given. 

Case II. Two sides and the included angle are given. 

Case III. Two sides and the angle opposite to one of them 
are given. 

Case IV. All three sides are given. 

53. Case I. Given one side and two angles. Let c, A, B 

be the given quantities. We may use the formulse 



(1) 



<7=180°- (A+B), 

tf sin A i e sin B 

a = — , o = 



sin Q sin C 

to compute (7, a, and b. 

The most reliable and convenient check is furnished by one 
of Mollweide's equations. (See Equations (6), Art. 51.) 

(2) a - =c^l(A-B) 

v J cos i 

The law of tangents may also serve as a check instead of (2), 
but it is not quite as convenient. In calculating the check, 
it is convenient to think of A as larger than B, so that A — B 
may be positive. If A is less than B, interchange A and B, 
and also a and b in equation (2). 

EXERCISE XXIV 

Example 1. Given c = 327.85, A = 110° 52'.9 } B = 40° 31'.7. Find C, 
a, and b. Check the results. 







|c = 


32; 


r.85 


(1) 


Given 


\ A = 


L10 


52'.9 


(2) 




A 


[b = 


40° 


31'.7 


(3) 




+ B = 


151° 


24'.6 


(4) 






C = 


28° 


35 . 1 


(5) 



TWO SIDES AND INCLUDED ANGLE 109 

Solution. 

log c = 2.51568 (6) 
log sin A = 9.97049-10 (7) * 

colog sin C = 0.32008 (9) f 

log a = 2.80625 (10) 

a = 640.10 (12) 

log c = 2.51568 (6) 

log sin B = 9.81280-10 (8) 

colog sin C = 0.32008 (9) 

log b = 2.64856 (11) 

b = 445.20 (13) 
Check. 

A - B- 70° 21 '.2 (14) logc = 2.51568 (6) 

\(A-B) = 35° 10'.6 (15) log sin \(A - B) = 9.76050-10 (17) 

|C=H°17'.7 (16) colog cos I (7 = 0.01366 (18) f 

log(« -b) = 2.28984 (19) 
a - b = 194.90 (12)-(13) 
a - ft = 194.91 (From (19)) 

Remarks. The numbers (1), (2), etc., indicate the order in which the 
separate results are written down and are meant to assist the student in 
understanding the arrangement of the computation. These numbers 
should not appear in the student's own work. 

Solve and check the following triangles. 

2. a = 467.00, A = 56° 28' .0, B = 69° 14'. 

3. a = 24.31, A = 45° 18', B = 22° 11'. 

4. a = 148.30, A = 37° 24'.0, C = 76° 48'.5. 

5. .1 =71 13' 30", B = 40° 34' 15", c = 236.54. 

6. a = 3.4356, A = 17° 43'.4, C = 60° 35'.7. 

7. .4 = 47° 13'. 2, B = 65° 24'. 5, a = 43.176. 

8. fi = 100 21' K>". C = 58° 17' 20", a = 31.656. 

9. a = 52.780, .1 = 37°41' 15", B = 7T 29' 40". 
10. .1 = 57 23' 12". C = 68° 15' 30", c = 832.56. 

54. Case II. Given two sides and the included angle. 
Let a, b. and C be the given parts. If a > 6, we use the for- 
mula 
(1) \(A + B) = W>-\C 

* Remember that sin 110° 52' .9 = sin (90° + 20° 52'.9) = cos 20° 52'.9. (Art. 43.) 
t Remember the rale for finding a cologarithm. (Art. 23.) 



110 SOLUTION OF OBLIQUE TRIANGLES 



to find ^(A 4- i?), and the law of tangents, 

a — b, isA, z>\ a — b 
a + b AK y a + b 



(2) tan \(A - B) = ^—1 t an J (J. + 5) = ^^coti (7, 



to find 1 (J. - 5). We then find J. and B from 

(3) ,4 = 1(^ + 2?) +K^-5), 

and apply the law of sines to find c, giving 

r a % « sin (7 

( 4 ) c== -^-J- 

As checks we use the relations 

(5) A + B+C =180°, 

and one of the Mollweide equations, in the form 

(6) a ~ h = sinl(vl-^) 

c cos J 

The law of sines furnishes another relation which might 
be used as a check instead of (6). But (6) is more reliable. 

If a < b, we interchange the letters a and b, and also A 
and B in all of the above formulae, to avoid the appearance 
of negative angles. 

EXERCISE XXV 

Example 1. Given a = 469.71, b = 264.37, C = 96° 57'. 6. Find A, B, 
and c and check the results. 

Solution. 

rC = 96°57'.6 (1) i C = 48° 28'.8 (6) 

Given a = 469.71 (2) " log coH C = 9.94711 - 10 (8) 

1 & = 264.37 (3) log (a - b) = 2.31247 (10) 

a + 6 = 734.08 (4) colog(a + 6) =7.13425-10 (11) 

a -b = 205.34 (5) log tan \ (A - B) = 9.39383 - 10 (12) 

!(^+5) = 41°31'.2 (7) 

log a = 2.67183 (18) \(A - E) = 13° 54'.6 (13) 

log sin C = 9.99679 -10 (19) , ^ = 55° 25'.8 (15) 

colog sin A = 0.08437 (20) B = 27° 36'.6 (16) 

logc = 2.75299 (21) C = 96° 57'!e (1) 

c - 566.23 



TWO SIDES AND AN OPPOSITE AX(iLK 



111 



Checks. 

log (a - 6) = 2.312 17 (10) log sin \(A - B) = 0.38001 - 10 (14) 

logc = 2.75200 (21) \ogcos\C = 0.82144- 10 (0) 

log^ = 0.55048 -10 (22) log fllllM-^ = 0.55047 - 10 (23) 
c cos i ( 

A + B + C = 180° O'.O (17) 

The checks consist in the result (17) and the close agreement of (22) 

and (23). 

Solve and check the following triangles : 



2. 


b = 472. 


c = 324, 


,1 = 78° 40'. 


3. 


a = 748. 


h = 37."-. 


C = 63° 35'. 5. 


4. 


a = 42.38, 


J = 35.00, 


C = 43° 14' 40". 


5. 


b = 0.941, 


c= 1.256, 


A = 35° 17' 28". 


6. 


a = 12.3460, 


6 = 5.7213. 


C = 65° 30' 10". 


7. 


a = 25.384, 


c = 52.025. 


B = 28° 32' 20". 


8. 


6 = 0.14367, 


c = 0.11412, 


.4 = 42° 14'.6. 


9 


a = V - 


& = 226.19, 


C = 50° 12' 54". 


O. 


5 = 1436.7, 


c = 1141.2, 


.4 = 42° 14' 35". 



55. Case III. Given two sides and the angle opposite to 
one of them. 

Geometric discussion. Let A, a, b be the given parts. We 
construct the angle XAY equal to the given angle A, and 
lay off AC. on AY, equal to the given side b. With C as 
center, and a radius equal to the given side «, we strike an 
arc. If this arc intersects AX in B and B\ one or both 
of the triangles ABC and AB' C may be solutions of the 
problem. 





Mj/ 


\a 


S\ ^/ 


V 


i ir—jr^B 


Fig. 


.",() 



The following cases may present themselves if A is an 



acute angle 



I. a< OP, i.e. a<bainA. (Fig. 48.) 

The triangle is impossible in this case, sine.' the arc of 
radius a. with C as center, will not intersect AX. 



112 



SOLUTION OF OBLIQUE TRIANGLES 



II. a =OP = b sin A. (Fig. 49.) 
The solution is the right triangle APC. 

III. b>a>bsinA. (Fig. 50.) 

There are two solutions in this case. The triangles ABC 
and AB'O both satisfy all of the requirements of the prob- 
lem. 

IV. a ^ b. (Fig. 51.) 

If a > 6, there is one solu- 
tion only ; namely, ABO. The 
Fig. 51 triangle AB' does not con- 

tain the given angle A, but its 
supplement. If a = b, ABC is isosceles, and AB' C reduces 
to a triangle of zero area. Thus we may say that there is 
one solution only if a > b. 

The following additional cases may present themselves if 
A is an obtuse angle. 






Fig. 52 

V. a < b. (Fig. 52.) 

The triangle is impossible if a < b and of zero area if a = b. 

VI. a > b. (Fig. 53.) 

There is one solution ; namely, the triangle AB C. The 
triangle AB' C does not contain the given angle A, but its 
supplement. 

Trigonometric discussion. How do the equations of trigo- 
nometry bring into evidence these various cases ? 

The law of sines gives 

b sin A 



(i) 



sin B — 



TWO SIDES AND AX OPPOSITE ANGLE 113 

If A, a, b are given, we may compute the right member of 
this equation. Suppose tirst that A is an acute angle. 

If the right member of (1) is greater than unity, the 
triangle is impossible, since the sine of an angle is never 
greater than unity. This is Case I of the geometric discus- 



es 
sion. 



If = 1, B must be a right angle. (Case II.) 

a 

If the given values of a, 6, and the acute angle A, are such 

that 

(2) bzinA <h 

a 

we can find not merely one angle to satisfy equation (1) but 
two snch angles, which are supplementary to each other, one 
acute and one obtuse. For if K is the acute angle which 
satislies equation (1), the obtuse angle 180° — i£ which has 
the same sine as K (See Art. 40), will also satisfy equa- 
tion (1). 

Thus we find in the first place, from (1), the two pos- 
sibilities 

(3) B = K (acute angle), B = 180° - K (obtuse angle). 

We must remember, however, that A+ B must be less than 
180°. If, as we have supposed, A is an acute angle, A+ K 
is certainly less than 180°. But A -f 180° - K is less than 
180° if and only if A < if; that is (see Fig. 50, where 
Z K= Z ABC), if and only if a< b. Only in this case 
then will there be two solutions. That is, we have two 
solutions if A is acute and if b > a > b sin A, in accordance 
with Case III of the geometric discussion. 

If A is acute, but if A + 180° - K > 180° ; that is, if 
A > K and therefore a > b, the obtuse angle solution 
180° — K for B becomes inadmissible and the problem has 
only one solution, in agreement with Case IV of the 'geo- 
metric discussion. 

If A is an obtuse angle, the obtuse angle solution 180° — K 
for B is never admissible, since a triangle can contain at 



114 SOLUTION OF OBLIQUE TRIANGLES 

most one obtuse angle. The acute angle solution B = K 
is admissible if and only if A + K is less than 180°. This 
distinction gives rise to the two remaining Cases, V and VI, 
of the geometric discussion. 

In the trigonometric solution of a numerical problem of 
this kind, it is essential to remember the following facts : 

1. If, on computing the sine of an angle, we find its value to 
be greater than unity, the triangle is impossible. 

2. If the sine of an angle is found to be a positive proper 
fraction, there are two possibilities for the corresponding angle. 
One of these angles is acute and the other, the supplementary 
angle, is obtuse. 

3. The sum of any two angles of a triangle must be less than 
180°. 

The sine of B having been found from the law of sines, as 
indicated above, it will become apparent from the correspond- 
ing values of B whether the number of solutions is 0, 1, or 2. 
If there is one solution, we find from 

A + B+ (7=180° 

and c from the law of sines. We may check by one of 
Mollweide's equations or by the law of tangents. If both 
values of B are admissible, we use each of them in succession, 
so as to find the remaining parts of the two triangles which 
are solutions of the problem. 

EXERCISE XXVI 

Example 1. Given A = 15° 32'.7, a = 103.21, b = 152.37. Find -the 
remaining parts of the triangle or triangles determined by these data. 

Solution. 

Formula : sin B = b sin A , C = 180° - (A + B), c = as ' mC . 
a sin A 

Check: ft- g== ™ in *Q8-^). 

cos \ C 

<A = 15° 32'.7 (1) (B = 23° 18'.4 B> = 156° 41.6 

Given a = 103.21 (2) Results C = 141° 8'.9 C'~ = 7° 45.7 
I b = 152.37 (3) i c = 241.58 c' = 52.01 



TWO SIDES AND AN OPPOSITE ANGLE 115 



Computation. 










log b = 2.18290 


(V 


loga= 2.01372 




(5) 


log- sin A = 9.42813 


(7) 


log sin C = 9.79748- 


10 


(15) 


cologa = 7.98628-10 


(6) 


cologsin .1 = 0.57187 




(17)* 


log sin/i = !».5!)7:U-10 


(8) 


Logc = 2.38307 




(18) 


B= 23°18'.4 


(9) 


c = 241.58 




(20) 


B' = 156°41'.6 


(10) 


log a = 2.01372 




(•>) 


B + A= 38°51'.l 


(11) 


log sin C = 9.13050- 


10 


(16) 


IV + A = 172 14'.3 


(12) 
(13) 


colog sin .1 = 0.57 1ST 




(17)* 


r = Ul 8'.9 


logc' = 1.71609 




(19) 


C" = 7°45'.7 


(li) 


c' = 52.01 




(21) 


Cftecfc 










B 


- X = 


7° 45'.7 (22) 






B' 


- .1 = 


141° 8'.9 (23) 






KB- 


--0=-- 


3° 52'.9 (24) 






l(B'- 


-n - 


70°34'.5 (25) 







logc = 2.38307 (18) log c' = 1.71609 (19) 

log sin \(B- A ) = s.83056-10 (26) log sin \(B'-A ) = 9.97455-10 (27) 

cologcos \ C- 0.47811 (28) colog cos £C" = 0.00100 (29) 

log (b - a) = 1.69174 (30) log (6 - a) =1.69164 (31) 

fr-a=49.17 (32) f b - a=49.16 (33)J 

ft-a=49.16 (34)| 

Example 2. Given A = 15°32 / .7 J a = 10.321, b = 152.37. Find the 
remaining parts of the triangle or triangles determined by these data. 

Solution . 

( A = 15° 32'. 7 log b = 2.18290 

Given a = 10.321 log sin A = 9.42813-10 

[ b = 152.37 cologa = 8.98628-10 

log sin B = 0.59731 
Since log sin B has the characteristic zero, sin B is greater than unity. 
Therefore, the triangle is impossible. 

Example 3. Given A = 15°32'.7, a = 107.38. I = 152.37. Find the 
remaining parts of the triangle or triangles determined by these data. 

Solution. 

[ A= 15°32'.7 \B= 14° 7'.2 

Given > a = 16*3 Results C = 150 20'. 1 

( 6= 152.37 [ c = 309.11 

* Obtained from (7). t Obtained from the logarithm above. 

| Obtained by subtraction from (2) and (3). 



116 



SOLUTION OF OBLIQUE TRIANGLES 



Computation. 

log 6 = 2.18290 
log sin A = 9.42813-10 
colog a = 7.77629-10 




log 

colog 


log a = 2.22371 

sin C = 9.69454-10 
sin A = 0.57187 


log sin B = 9.38732-10 

B = 14° 7'.2 

£'=165°52'.8 

B + A = 29° 39'.9 

B' +A =181°25'.5* 

C = 150°20'.l 




log c = 2.49012 
c = 309.11 


Check. 

A -B= 1° 25'.5 
\{A-B)= 0°42\8 


log 


logc = 2.49012 
sin \(A -B)= 8.09516-10 
colog cos I C = 0.59180 


\ C = 75° 10M 
a-b = 15.01 




log (a -6) = 1.17708 
a - 6=15.03 



Find out whether the triangles corresponding to the following data 
are possible, how many solutions there are, and what are the values of 
the missing parts. 

4. a = 98, b = 100, 

5. a = 767, 6 = 242, 

6. a = 3541, 6 = 4017, 

7. a = 67.53, 6 = 56.82, 

8. a = 9.4672, c = 14.433, .4 = 11° 14'.3. 

9. a = 413.28, 6 = 378.19, B = 50° 16' 25". 
10. a = 345.46, 6 = 531.75, ^ = 26° 47' 32". 



A = 120°. 
A = 36° 53' 2". 
^ = 61° 27'. 
A = 77° 14' 19". 



56. Case IV. Given the three sides of the triangle. We 

use the formulse 



8 = 2 ( a + ^ + 



, r = \^ — 



— a)(s — £>)(« — c) 



A. A. A» 

tan l J. = , tan^i? = , tan I = 



s — a 



s-b 



— c 



and the check 



A + B+ (7=180°. 



* Therefore B' is inadmissible. There is only one solution, as might have 
been foreseen, since a>6. 



GIVEN THE THREE SIDES 117 

EXERCISE XXVII 

Example L. Given a = 34.278, 6 = 25.691, c = 30.175. Find the 
angles I. B, C. 

Solution. 

[a=34.278 f -1- 75 13'.0 

Given ' 6=25.691 Results B= 16 26'.6 

1 c= 30.175 [ C= 58° 20'.2 
_ = 90.144 Cluck: A+B+C=179 59'.8 

s= 45.072 cologs = 8.34609 -10 

s-a=10.794 log (s --a) = 1.03318 

.,_ A = 19.381 log (s- 6) = 1.28737 

. s--c = H j " 1og(s- c) = 1.17310 

2 s = 110.144 * log r 2 = 1 .83974 f 



logr=0.91987 logr=0.91987 log r= 0.91987 J 

log(*-q) = 1.03318 1 lo g (.*-/>) = 1.28737 . j logQ-c) = 1.17310 
log tau A A = 9.88069 - 10|log tan .] B = 9.63250 - lOlog tan .\ C = 9.74677 - 1 
|4=37°36'.5 J5=23 13'.3 |C=29°10'.l 

Remark. In this problem some time may be saved by writing log r on 
the lower margin of a slip of paper and placing it above log (s — a) to 
find log tan ] . 1 . above log (* — h) to find log tan J B, and above log (s — c) 
to find log tan \ C. A similar device is often useful in similar cases. 
Most computers also Bave time by omitting the — 10 attached to logarithms 
with negative characteristics. This omission can never give rise to 
serious misunderstanding. 

Find the angles of the triangles whose sides have the following values : 
2. a = 79.3. b = 94.2, c = 66.9. 



3. 


a = ii., B5, 


6 = 0.850, 


c = 0.633. 


4 


a = 312. 


6 = 423. 


c = 342. 


5. 


a = 25.17. 


6 = 34.06, 


c = 22.17. 


6. 


a = 931 16, 


6= 17653M. 


C = 9576S. 


7. 


a = 12.653, 


6 = 17.213, 


r = 23.1o6 



If only one of the three angles is to be calculated, it may be more con- 
venient to make use of the formulae for sin \ .1 or cos £ A, which may be 
found from the indications given in Example 4, Exercise XXII. 

♦Obtained by adding 8, 8 — a, 8 — b, 8 — c t as a check on the additions and 
subtractions required to find these qoantiti 

t Obtained by adding the four logarithms above logr 2 , since 
r2 _ (s — ti)(s — b)(s — n) t 

8 

t Obtained bj taking one half of log 



118 SOLUTION OF OBLIQUE TRIANGLES 

57. Problems in heights and distances, plane surveying, and 
plane sailing. Some of the following problems are direct appli- 
cations of the methods which have just been explained. In 
others, it will be necessary to consider several triangles in 
succession or simultaneously. 

It will always be advisable to draw a figure, approximately 
to scale, to denote the known as well as the unknown sides 
and angles of the figure by properly chosen letters, and to 
write down the formulae to be used in their general form, 
leaving the substitution of the numerical values to the last. 
Much blundering and much unnecessary work may be avoided 
by adopting this plan. 

The student should use his judgment in regard to the 
number of decimal places used in the numerical part of the 
work. Use three-place tables whenever possible. Many of 
the problems may be solved, wholly or in part, by the slide- 
rule. 

EXERCISE XXVIII 

1. In order to find the height of a tower, the angles of elevation of 
its top are measured from two stations, A and B, in the same horizontal 
line with its base, A being the more remote station. If the angles of 
elevation of the tower from A and B are 32° and 65° respectively, and if 
the distance AB is 500 feet, find the height of the tower. 

2. Find the distances from the two stations of Ex. 1 to the foot of the 
tower. 

3. If the angles of elevation of the tower from A and B, in Ex. 1, 
are L and M respectively, and if the distance AB is equal to d feet, show 
that the height of the tower is 

, _ r/ sin Z sin M 
1 ~ sin (M - L) ' 

4. An obstacle (a house) was found to interfere with the running of 
a straight line from A in the direction AB. (See Fig. 54.) An angle 

ABE was turned at B, equal to 123°, and 
the distance BE was measured equal to 
150 feet. The angle BEC was made equal 
to 63°. How long must the distance EC 
be, and what angle must be turned at C, in 
order that CD may be the prolongation of 
AB1 




APPLICATIONS [NVOLVTNG OBLIQUE TRIANGLES 111) 

5. How may angles at I) ami E be chosen, in Ex. I, so as to avoid 

all computation? 

6. Two straight railroad tracks intersect at an angle of 75°. What 
will be the distance, at the end of 20 minutes, between two trains which 
start from the crossing at the same instant, their speeds being 30 and 10 
miles per hour respectively? 

7. Each oi two battleships passing each other fired a salute. A 
person on shore observed that the interval between the flash and the 
report of the gun was 1 seconds for one ship and 6 seconds for the 
other. The angle, at his eve, subtended by the two ships, just as 
the salute was tired, was 55 . The velocity of sound is about 1110 feet 

- ond. Find the distance between the ships. 

8. Two light houses are 2.789 miles apart, and a certain rock is known 
to be 4.325 miles from one of them. The angle subtended by the two 
lighthouses at the rock is 1(5° 13'. How far is the rock from the other 
lighthouse? How many solutions are there to this problem? Can we 
make a choice between these solutions if we know which of the two light- 
houses is nearer to the rock? 

9. Find the radius of the largest cylindrical gas tank which can be 
constructed on a triangular lot whose sides measure 73, 82, 91 feet respec- 
tively, and locate the center of its circular base. 

10. An observer measures the angle of elevation of a cloud due south 
of him at the moment when the sun also is due south '(at apparent noon). 
The angle of elevation of the sun was 65°, that of the cloud 75°. If the 
shadow of the cloud falls 550 feet north of the observer, how high is the 
cloud? 

11. At p.m. two lights, known to be 8 miles apart, are observed to 
be due east from a certain vessel. At 10 p.m. one of these lights bears 
N.E. and the other X.X.E. If the course of the ship was due south, what 

was its r;r 

12. A tower is situated on top of a conical hill whose sides make an 
angle of 15° with the horizontal plane. At a distance of 120 feet from 
the foot of the tower (tie- distance being measured along the slope) the 
tower subtends an augle of -Jo . Find the height of the tower. 

13. If. in Ex. 12. the side of the hill makes an angle / with the hori- 
zontal plane, and if the angle subtended by the tower, at a distance of <1 
feet from its foot, i- .1. -how that the height of the tower is 

'/ sin .1 



h = 



I + /) 



120 



SOLUTION 'OF OBLIQUE TRIANGLES 




14. A tower 54 feet high, situated on top of a conical hill, subtends an 
angle of 15° 30' at a point 120 feet from the foot of the tower (the 
distance being measured along the slope). What angle does the side of 
the hill make with the horizontal plane? 

15. To find the slope of a railroad embankment, one end of a pole 12 
feet long was placed on the level ground 6 feet from the foot of the em- 
bankment, and the other end was found to fall at a point 7.5 feet up its 
face. What angle does the embankment make with a horizontal plane? 

Remark. A transit cannot conveniently be used to measure an angle 
formed by two walls, the angle formed by an embankment or buttress 
with a horizontal plane, etc. In such cases, as in this example, it is 
more convenient to measure distances and deter- 
mine the angles by calculation. 

16. Two capes, A and B (Fig. 55), were ob- 
served from a ship at sea ; one of them bore 
N.N.E. and the other N.W. It was found from 
the chart that the second cape bore W. by N. 
from the first and was 25.3 miles distant from it. 
What was the distance of the ship from each of 
the two capes ? 

17. A battleship leaves port A, on a due easterly course, at the rate of 
16 miles per hour. A dispatch boat starts from B at the same moment. 
The port B bears S.S.W. of port .4 and is 25 

miles distant from it. If the dispatch boat has a 
rate of 22 miles per hour, what should be the 
direction of its course so that it may meet the 
battleship, if neither ship alters its rate or course ? 
At what time will they meet ? 

Hint. In Fig. 56 we have AC = lQt, BC = 
22 t, if t denotes the time (in hours) which passes 
between the time of sailing and the moment of meeting, and if C repre- 
sents the place of meeting. 

18. The angle of elevation of the top of a tower, at a point in the 
same horizontal plane with its base, is equal to A. At a point h feet 
directly above the first the angle of depression of the foot of the tower 

was found to be equal to B. Prove that 
the height of the tower is equal to h tan A 
A 1 B__ C ot 5. 

19. A valley has the cross section shown in 
Fig. 57, the angles L and M and the distance 
AB = I having been obtained by a survey. It 
is planned to connect the points A and B by a 




Fig. 56 




APPLICATIONS tNVOLVING OBLIQUE TRIANGLES 121 

Horizontal bridge, supported by a pier at C. How high must this pier 
be made? 

20. Show that the area of a quadrilateral is \ ill' sin .1. if rf and <l' 
are the lengths of its diagonals and if A is one of the angles which the 
diagonals make with each other. 

21. Two sides o( a parallelogram are 3.41 and 2.60 feet long and the 
length of one diagonal is 1.58 feet. Find the length of the other diagonal. 

22. The sides of a tield . 1 B( 'D are : . I B=57 feet, BC=43 feet, CD=45 
feet. DA = 47 feet: and the distance from .1 to (' is 50 feet. Find the 
area of the tield. 

23. Two streets intersect at an angle of 75°. The corner lot has 
frontages of 150 feet and 115 feet on the two streets, and the remaining 
two boundary lines of the lot are perpendicular to the two streets. What 
is their length, and what is the area of the lot? 

24. In order to measure the distance between two pumping stations, 
A and B, in Lake Michigan, a base line CD = 17.7 chains was measured 
along the shore. (See Fig. 58.) The follow- b 
ing angles were measured : 

ACD = Ci = 132°29', 
,1 CB = 82° 20', 
< DA =2>i = 45°59', 
QDB = Z> 2 '=124°48. 
Compute the distance A B. 

25. Devise a plan for finding the distance between any two inacces- 
sible points. A and B, in the same horizontal plane if two points, C and 

D, can be found in the same plane, from 
both of which A and B are visible. 

26. Devise a plan for finding the dis- 
tance between two inaccessible points, .1 
and 11. if both are visible from only one ac- 
cessible point C. 

Hint. In Fig. 50, select a point D from 
which A and C are visible, and a point E from which B and (' are 
visible. Measure CD, CE, and the angles Ik C%, <"•_>. C a , E. 

27. To compute the distance between two accessible points, A and /;. 
if no point can be found from which both .1 and II can be seen. (For 
instance, if .1 and P> are points on opposite sides of an inaccessible 
mountain.) Take ;i point C from which A may be seen and a point D 
from which II i- visible. If C is visible from I), measure the angles 
A CD and CDB and the distances AC, ('Ik DB. show how to calcu- 
late the distance All from these data. 





Fig. B9 



122 



SOLUTION OF OBLIQUE TRIANGLES 




Fig. GO 



28. If the points A and B of Ex. 27 are inaccessible, the distances 
A C and BD cannot be found by direct measurement. In such a case 

(see Fig. 60), select points C, D, E, F so 
that A , C, E shall be visible from D, and 
D, F, B from E. Measure the angles 
C, Di, Z>2, Ei, Ei, F, and the distances 
CD, DE, and EF. Show how to find the 
distance AB from these measurements. 

29. A tower is situated on top of a 
conical hill as in Ex. 12. Two points 
A and A' are chosen on the side of the 
hill at distances d = 43 feet and d' = 98 
feet respectively from the top, the point ,4' being the lower one and 
the distances d and d' being measured along 
the slope. The angles subtended by the 
tower at A and A' were A = 42°, A' = 23° re- 
spectively. Find the height of the tower and 
the angle of inclination of the side of the hill. 

30. Two observers, A and B, are 3 miles 
apart, A being due west of B and in the 
same horizontal plane. Both observers meas- 
ure, at the same instant, the bearing and 
angle of elevation of a balloon. A finds that 

the balloon bears N. 47° E. and that its angle of elevation is 23°. B finds 
that the balloon bears N. 35° W. and that its angle of elevation is 42°. 
Find the height of the balloon above the horizontal plane of A and B. 

31. To find the horizontal distance AD and the vertical distance DC 
from A to an inaccessible point C (Fig. 62) when it is not convenient 

to measure a base line in the 
same vertical plane with C. 

Measure a horizontal base 
line AB, k feet long, in any 
direction through A . Let D be 
the foot of the perpendicular 
from C to the horizontal plane, 
MN, of AB. Measure the hori- 
zontal angles BAD = A and 
ABD = B, and the vertical an- 
gles DA C = A' and DBC = B>. 




Fig. 61 




Fig. (52 



Show that 



AD 



k sin B 
sin {A + B) ' 
k sin B tan A 



BD = 



k sin A 



sin (A +B) 
k sin A tan B' 



sin (A + B) sin {A + B) 



APPLICATIONS INVOLVING OBLIQUE TRIANGLES 123 



The two formulae for h should give the same result in any numerical 
problem of this kind. Lack of agreement indicates inaccuracy in one 
of the observed angles or in the computation. 

32. Two points, .1 and />'. in the same horizontal plane and separated 
by a ridge, are to be connected by a straight level tunnel. In order to 
find the distance between them, the surveyors measured the inclined 
angle, BCA, subtended by AB from the top C of a neighboring hill, 

whose height. CD = /i. above the plane of A and B is known. They 
also measured the angles o\ depression of .1 and 5 from C. Devise a 
method for computing the length of the tunnel AB. 

33. Prove, by means of the trigonometric formulae, that the angles of 
a triangle can be found when the ratios of the three sides of the triangle 
are given, even if the absolute values of the three sides are not known. 
Prove the same fact by geometry. 

34. Let . 1. B. C be three points of a horizontal line, whose mutual 
distances, AB- ft, AC - c, BC = b + c, are known. (See Fig. 63.) To 
find the horizontal distances, />. 7, r, of an inaccessible point, E, from 
these three points, and the height h, it suffices to measure the three angles 



of elevation. .1. B. (\ 
A. />. C respectively. 
For the figure gives 


of E from 


(1) p — h cot .1. q : 
r — h cot C, 
and also 


= h cot B, 


,f = h- + pa - 2 hp cos BA D. 

r- = c 2 + p 2 + 2 rp cos B. 1 D ; 
whence 


})' 2 4- p' 2 — 7- r' 2 — r' 2 — p 2 




Fig. 63 



Substitute the values of p, 7. r from (1) and solve the resulting equa- 
tion for 4 s . We find 



(2) 



h 2 = 



hc(h -j- c) 



(cot - C - cot 2 A )h + (cot a B - cot- < I )c 
After computing h from (2), p. 7. and r may be found from (1). 

35. Given the mutual distances of three points I. />'. C,to find the 
distances Ah. I',I>. CD from a fourth point T) in the plane of .I/;*" to 
each of the given three points, when the angles art- given which the lines 
.!/;. BC, CA Bubtend at D. 

This problem, usually called PotJienot's problem, may be solved as f.,1- 



124 



SOLUTION OF OBLIQUE TRIANGLES 



In Fig. 64, let 

BC = a, CA =6, AB = c 

be the given mutual distances of A, B, and C. Of course the angles of 
the triangle ABC may then also be regarded as known. 

Let Z CDB = L, Z CDA = M, Z BDA = N 

be the angles subtended by the sides of the triangle 
from D. These angles we regard as known by meas- 
urement and, of course, N = M -f L. 

The problem of finding the distances AD, BD, CD 
may evidently be regarded as solved, if we can find 
the two angles 

ZCAD = P and Z CBD = Q. 

We shall show how to compute these angles. 

Applying the law of sines to the triangles A CD 
and BCD, we find 

b sin P _ a sin Q 
sin M sin L ' 
sin P a sin M 




(1) 

whence 



CD = 



sin Q b sin L 
Hence, by the theory of proportion, 

sin P — sin Q a sin M 



b sin L 



sin P + sin Q a sin M + b sin Z 
which gives (Equations (8), Art. 49), 



(2) 



tani(P-Q) _ 
tani^+Q) j 



1- 



5 sin X 
o sin M 



b sin Z 



a sin ili" 
On the other hand we have 

p + Q + C + N= 360°, 
and therefore 

(3) 1{P + Q) = 180° -KC+iV). 

Since the angles C and N are known, (3) gives the value of \(P + Q). 
If this value be substituted in (2), we may obtain from (2) the value of 
l(P - Q). From |(P + Q) and \(P - Q), we find P and Q themselves 
by addition and subtraction. The distance CD may then be computed 
by (1) in two ways, providing a check for the correctness of the work. 

Remark 1. The problem obviously becomes indeterminate, if the 
point D should happen to be on the circumference of the circle deter- 
mined by the three points A, B, C. For, as the point D moves along 
this circumference, the angles L and M do not change, so that the posi- 
tion of the point D on the circumference is not determined by the value 



APPLICATIONS ENVOLVTNG OBLIQUE TRIANGLES 125 

of those angles. It is evident, then, thai if the point D is close to the 
circumference of the circle circumscribed about the triangle ABC t .ita 
position cannot be determined by this method with any considerable 
degree of accuracy. 

Remark 2, The name Pothenot's problem cannot be justified his- 
torically. A complete solution oi this problem was given by Snellius 
in his Doctrinal triangulorum canonicce libri quatuor, which appeared in 
1627, almost seventy years before Pothenot presented his solution of the 
same problem to the Paris Academy of Science. 

36. Show that the problem of Pothenot may also be solved by draw- 
ing a circle through the three points .1. B, !>. and by making- use of the 
triangle ABE, when 1 E is the second point of intersection of CD with 
this circle. 

37. In surveying a harbor, a submerged rock was located, for chart- 
ing purposes, by sighting three known objects .1. B, C on land from a 
boat immediately above the rock. The known distances were BC = a 
= 312 feet. CI = h — 520 feet, and the angle C was known to be 65° 27'. 
The angles obtained by observation were L = CDB = 23° 25' * and 
M = CDA = 32* 52'. Find and check the distance from the rock to C 
and the angle which this line makes with the side AC of the known 
triangle. 

In Exs. 38 to 45, we use the notations of Chapter VII; A, B, C for 

the angles, a. />. c for the sides, 2 s for the perimeter, rand R for the 
radii of the inscribed and circumscribed circle respectively, and S for 
the area of the triangle. Show how to find all of the sides and angles 
of the triangles determined by the following data. 

38. a — h. c. A — B are given. 42. r, -1, B are given. 

39. n - 1,. c, A — V> are given. 43. S, A, B are given. 

40. /.'. n, h are given. 44. S. a. h are given. 

41. R, .1. /; are given. 45. s, R, a are given. 

46. A furnace maker receives an order for a number of furnaces, some 
40 inches and some 42 inches in diameter. These furnaces are to be fitted 
on the outside with an iron casting whose inside length, measured along 
the arc. is 20 inches. In order to avoid the necessity of making two dif- 
ferent castings, the manufacturer considers the possibility of making a 
Bingle . 'lit exactly* furnace 41 inches in diameter but. of course, 

not fitting exactly either of the sizes ordered. His experience tells him 
that such a casting will serve the purpose, if no point of its inner surface 
is more than a quarter of an inch from the outer surface of the furnace 
after being placed in position. Will it be necessary to make separate 
gS for tin- two different sizes? 



126 



SOLUTION OF OBLIQUE TRIANGLES 



58. Displacements, velocities, and forces. If a body is 
transported from one place in the plane to another, and we 
wish to describe its change of position or displacement, it is 

clearly not sufficient to state how 
far the body has been moved. We 
must also include in our descrip- 
tion a statement concerning the di- 
rection of the displacement. 



N 



W- 



s 

Fig. 65 



Thus, the displacement from to P 
(Fig. 65) may be described completely by 
stating its magnitude (the length of the 
line-segment OP), and its direction (either 
the bearing NOP of the line OP or the 
angle EOP or some other angle which 
fixes the direction of OP). 



If the displacement is in a horizontal plane and the direc- 
tions from to iV", S, E, TTin Fig. 6o represent north, south, 
east, and west respectively, the projections OP' and OP" of 
OP on OE and ON are called the easterly and northerly 
components of the displacement. If the displacement is not 
horizontal, we define, in a similar manner, its horizontal and 
vertical components. 

Let us suppose that a point M is displaced from to P. 
(See Fig. 66.) The displacement may be represented in 
magnitude and direction by the directed line-segment OP. 
(The arrowhead indicates that the displace- R 

ment is from toward _P, and not from P 
toward 0.) Let OQ represent a second 
displacement. If the point M, originally 
at 0, be made to undergo both of these dis- 
placements in succession (in either order), 
it will ultimately arrive at R, where R is the fourth vertex 
of the parallelogram determined by OP and OQ. For this 
reason, the displacement OR is said to be the resultant of 
the displacements OP and OQ. 

We may think of the two displacements OP and OQ as 
taking place simultaneously. An instance of this sort is 




Fig. 66 



DISPLACEMENTS, VELOCITIES, AND FORCES 127 

furnished by a passenger shifting his position on board of a 
moving- ship. His total displacement, in space, is the result- 
ant of that which is due to the motion of the ship and of 
that caused by his own muscular efforts. 

The velocity of a train, of a bullet, or of any uniformly 
moving object, is measured by the distance which it describes 
in a unit of time, that is, by a displacement. Therefore a 
velocity, like a displacement, has direction as well as magni- 
tude. The case of a passenger's stroll on board a moving 
ship suffices to illustrate the phrase: resultant of two velocities. 
Since the velocity of a uniformly moving body is its dis- 
placement in a unit of time, the resultant of two velocities is 
found by the same method as a resultant of two displace- 
ments, i.e. by the parallelogram construction illustrated in 
Fig. 66. 

It is one of the fundamental facts of mechanics (proved 
by countless experiments) that two forces acting upon the 
same material point combine into a single resultant force accord- 
ing to this same parallelogram laiv. This fact is generally 
known as the law of the parallelogram of forces. 

Owing to the fact that displacements, velocities, and forces 
are directed quantities which combine accdrding to the 
parallelogram law, these three classes of things have many 
properties in common. There are many other instances of 
quantities of this same character and, on account of their 
importance, they have received a special name. 

Directed quantities, which combine in accordance with the 
parallelogram law, are called vectors.* 

By a proper choice of the units, every vector may be 
represented by a directed line-segment or, what amounts to 
the same thing, by a displacement. 

Thus, the line-segments of Fig. 00 may be interpreted as forces, if the 
directions of these line-segments coincide with the directions of the forces, 
and if each segment Lb made to contain as many length units as there are 
force units in the corresponding force. 

* From the Latin vector, meaning one who carries or conv< 



128 SOLUTION OF OBLIQUE TRIANGLES 



EXERCISE XXIX 

1. A steamer is moving N.N.E. with a velocity of 16 miles per 
hour. Find the northerly and easterly components of its velocity. 

2. A horizontal force of 10 lb. and a vertical force of 24 lb. are acting 
simultaneously on a point. Find the magnitude and direction of the 
resultant. 

3. A schooner is sailing due west at the rate of 6 miles per hour. A 
sailor is crossing the deck, from south to north, at the rate of 3 miles per 
hour. What is the magnitude and direction of his velocity in space ? 

4. A force of 250 lb. is acting on a body in a direction which makes 
an angle of 17° with the horizontal plane. How much of this force tends 
to lift the body, and what part of it tends to move the body in a hor- 
izontal plane ? 

5. Two forces, of magnitudes 350 lb. and 510 lb., respectively, act 
upon the same point, in directions which make an angle of 35° with each 
other. Find the magnitude of the resultant, and the angles which it 
makes with each of the component forces. 

6. A force of 216 lb. is resolved into two components, which make 
angles of 27° and 32° respectively, with the direction of the original 
force. Find the magnitude of each component. 

7. A man wishes to reach a point on the opposite side of the river, 
250 yards upstream. The velocity of the current is 2.5 miles per hour 
and the width of the river is 300 yards. If the man's rate of rowing 
(in still water) is 4 miles per hour, in what direction must he point the 
head of the boat in order that his course may be a straight line ? 

59. Reflection and refraction of light. The path of light, 
in a homogeneous medium like air, is rectilinear. But if a 
ray of light meets the polished surface of a sheet of metal or 
glass, its direction is changed in accordance with the law 
that the angle of incidence is equal to the angle 
of reflection. 

This law is illustrated in Fig. 67, where 
the ray A strikes the reflecting surface of 
the mirror at and is reflected in the direc- 
tion OB. The line ON, perpendicular to 
the reflecting" surface at 0, is called its nor- 

Fir 67 

mal. The angle NOA, or i, is called the 
angle of incidence, and the angle NOB, or r, is the angle of 
reflection. According to the law of reflection of light (veri- 





REFLECTION AND REFRACTION OF LIGHT 129 

tied by thousands of experiments) these two angles are 
equal. 

When a ray of light AO, after passing through air, meets 
the bounding surface of a second transparent medium like 
glass (see Fig. OS), only a part of the 
Light is reflected. Another portion 
of the light enters the second medium 
and continues on its way in a path 
OB, which makes a certain angle with 
the direction AO of the original ray. 

If jVj-V' is the perpendicular or nor- 
mal to the bounding surface at 0, the 

i at/-> t • • 11 j i.i 7 • Fig. 68 

angle xY OA, or ?, is called the angle 

of incidence and N'OB, or r, is the angle of refraction. 

As a result of numerous experiments, it has been found 

that the quotient 

sin i 

sin r 

will have the same value, for a given kind of glass, for all 
different values of i. In other words, as the angle of inci- 
dence changes the angle of refraction also changes, but in 
such a way as to leave the quotient 

/ n sin i 

(1) - — = n 

sin r 

unchanged. This quotient n is called the index of refraction 
of the glass with respect to air. Its value is different for 
different kinds of glass. For ordinary crown glass n is 
about equal to 1.5. 

If the ray of light again emerges into air, as indicated in 
Fig. 68, after having passed through a sheet of glass with 
exactly parallel sides, careful measurements show that the 
ray BC is parallel to the original ray AO. In other words, 
the direction of a ray of light is not changed by passing 
through a sheet of plate glass whose two faces are exactly 
parallel to each other. Therefore : the index of refraction of 



130 



SOLUTION OF OBLIQUE TRIANGLES 



air with respect to glass is the reciprocal of the index of refrac- 
tion of glass with respect to air. 

The law of the refraction of light was first discovered by 
Snellius in 1618. The simple formula (1) was given by 
Descartes in 1637. 



EXERCISE XXX 

1. A light is placed on the line perpendicular to a plane circular 
mirror at its middle point. The distance from the light to the mirror is 
15.76 inches, and the mirror is 8.32 inches in diameter. Consider two 
rays which strike the mirror in the extremities of one of its diameters. 
What angle will they make with each other after reflection ? 

2. In an experiment, a source of light is to be placed at A, a mirror 
at B, and a photographic plate at C. The distances are: AB = 5.367 
meters, BC = 6.329 meters, CA — 7.361 meters. What angle must the 
mirror at B make with the line A B so that the light, reflected at B, may 
pass through C ? 

3. Two billiard balls, A and B, have been placed at distances a and b 
inches respectively from the same cushion. The line joining them 
makes an angle L with the cushion. Let K be the angle at which the 
first ball must strike the cushion, so as to hit the second after rebound- 
ing. Show that 

tan K = ° tan L. 
b — a 

Remark. Billiard balls not endowed with a lateral rotation (without 
" English ") rebound in accordance with the law of reflected light. 

4. Find the index of refraction from the following observations. 

(Wullner) 



Angles of incidence, i . . . 


40° 


60° 


80° 


Angles of refraction, r . . . 


24° 24' 


33° 38' 


' 38° 57' 



The three values obtained for n will disagree slightly owing to inac- 
curacies in the measurements. 

5. A ray of light strikes a plate of crown glass at an angle of inci- 
dence of 37°. Find the angle between the reflected and the refracted 
ray, if the index of refraction is 1.559. 



REFLECTION AND REFRACTION OF LIGHT 



131 




Fig. 



6. A ray of light .1 BCD passes through a 

- ;>ri>m whose cross section (see Fig. 69) 
is an equilateral triangle. If the index of 
refraction is 1.559, what must he the angle 
of incidence in order that the path of the 
light in the prism may he parallel to one 
of its face-? What angle will the ray CD make with its original direc- 
tion AB. after emerging from the prism? 

7. A ray of light. ABC. etc., enters a glass prism, 
whose cross section is an isosceles right triangle and 
whose index of refraction is 1.5, at the point 11 
(Fig. 70), at right angles to the face of the prism. 
At C no part of the ray can be refracted (Why ?). 
and all of it is reflected in the direction CE. Such 
a prism is called a total reflecting prism. 




Fig. 70 



THE GREEK ALPHABET 



a, 


A 


Alpha 


v, 


N 


Nu 


A 


B 


Beta 


fc 


*— * 
*— * 


Xi 


7, 


r 


Gamma 


o, 





Omicron 


s, 


A 


Delta 


w» 


n 


Pi 


e, 


E 


Epsilon 


Pi 


p 


Rho 


& 


Z 


Zeta 


(T, ? 


, 2 


Sigma 


?7» 


H 


Eta 


T, 


T 


Tau 


0, / 


>, © 


Theta 


y, 


T 


Upsilon 


i, 


I 


Iota 


+. 


<S> 


Phi 


*, 


K 


Kappa 


X* 


X 


Chi 


X, 


A 


Lambda 


* 


^ 


Psi 


V» 


M 


Mu 


G), 


n 


Omega 



132 



PART TWO 

PROPERTIES OF THE TRIGONOMETRIC 
FUNCTIONS 

CHAPTER IX 

THE GENERAL ANGLE AND ITS TRIGONOMETRIC 
FUNCTIONS 

60. The notion of the general angle. In elementary geome- 
try we usually think of an angle as ready-made. We there 
think of two lines as given and understand by the angle be- 
tween them a measure of their difference of direction. But 
many reasons urge us to oppose to this static idea what might 
be called the dynamic concept of angle, which presents an 
angle, not as the ready-made difference of direction between 
two fixed lines, but as something which is generated by the 
rotation of a straight line around a fixed point as pivot. 

Thus, for instance, we shall say that the minute hand of a clock 
describes, or generates, an angle of 90° in fifteen minutes, an angle of 
360 in one hour, an angle of 1890° in five hours and a quarter. Although 
the minute hand points to the same place on the face of the clock after 
any number of complete revolutions, \xe are not likely to make the 
error of ignoring these complete revolutions.* If we did, we should be 
ignoring the distinction between 1 o'clock. 2 o'clock, 3 o'clock, etc. If 
we were to say that an angle of 360 is the same as one of 0°, or that an 
angle of 450 is equal to one of !»<> . we should be committing the same 
error. 

We see that, while an angle in the sense of elementary 
geometry can never be greater than 180°, our now concept 
of angle permits us to speak of angles of any magnitude. 

* It is the purpose of the hour hand to record the number of complete 
revolutions. 

133 



134 .FUNCTIONS OF THE GENERAL ANGLE 

Our notions will be enriched in still another way, if we 
adopt the dynamic instead of the static concept of angle. 
The line, whose rotation generates the angle, may revolve 
in either of two opposite directions, clockwise or counter- 
clockwise; and we must distinguish between these two kinds 
of rotation, just as we distinguish between two motions in 
opposite directions on a straight line. This distinction may 
be made by ascribing to every angle, not merely a magnitude, 
but also a sign depending upon the direction of the rotation 
by which the angle is generated. 

It is customary to speak of counterclockwise rotations as 
positive, and of clockwise rotations as negative. 

The reason for this convention * will appear later (Art. 63). 

EXERCISE XXXI 

Using a protractor, combine the following angles graphically and 
check the results arithmetically. 

1. 30° + 60°, 50° - 30°, 30° - 60°, 50° + ( - 30°), 30° + ( - 60°). 

2. 25° + 15° - 35°, 135°- (- 25°)+ 150°. 

3. 225° + 345° - 185°, 30° + (3 x 15°). 

4. What angle does the minute hand of a clock describe in 3 hours 
and 25 minutes? in 5 hours and 13 minutes? 

5. Suppose that the dial of a clock is transparent so that it may 
be read from both sides. Each of two persons, stationed on opposite 
sides of the dial, observes the motion of the minute hand for fifteen 
minutes. Upon comparing notes, they find that they do not agree in 
regard to the angle described by the minute hand during this period of 
time. In what respect do they differ? 

6. What is the magnitude of the angle described by a spoke of a 
carriage wheel, 3 feet in diameter, when the carriage travels a distance 
of 500 feet? 

Note. Think of the wheel as if it were turning on the axle while the 
carriage is standing still. 

* The word convention is here used in a special sense, meaning an arbitrary 
agreement. 




INITIAL AXD TERMINAL SIDE 135 

7. The earth describes an approximately circular orbil about the 

sun as center in 365 days. What angle will the line joining the sun to 
the earth (the earth's null us rector) describe in 415 days? 

8. Two wheels. .1 and B, are joined by a belt as in Fig. 71. The 

diameter of -1 is twice that of />'. and A is moving in counterclockwise 
direction. What angle will a spoke of 
B describe while A rotates through an 
angle oi 300 ? 

9. If the two wheels of Ex. 8 are 
joined by a crossed belt, what angle will 
a spoke of /> describe when .1 rotates 
through an angle of 300°? 

10. If n wheels are connected by gears. 
what kind of a number must n be in order that the first and last wheel 
may rotate in the same direction? in opposite directions? 

61. Initial and terminal side. Standard position of an 

angle. Our new concept of an angle, as a measure for the 
, c amount of rotation of a line, leads us 
to distinguish between the initial and 
terminal sides of an angle. 

Thus, in Fig. 72, we have three angles 
whose senses of rotation are indicated by 
curved arrows. The angles A 0\B and CO2D 
p IG 72 are positive, for the rotation is counter- 

clockwise. Angle E0 3 F is negative. The 
initial sides of these angles are AOi, CO?, EOz, and their terminal sides 
are BO v DO* and FOi respectively. 

If an angle is thought of as generated by the rotation of a 
straight line, the initial and final positions of this line are called 
the initial and terminal sides of the angle respectively. 

If we wish to compare two parallel directed line-segments 
in regard to magnitude and sign, we usually think of one of 
them as being moved, until its initial point coincides with the 
initial point of the other. In the same way. in order to com- 
pare two angles we usually place them so that their vortices 
and initial sides shall coincide. 

It is customary, for purposes of comparison, to place all 





136 FUNCTIONS OF THE GENERAL ANGLE 

angles in such a position that their 
initial sides are on a horizontal line 
and pointed toward the right. An 
angle placed in this way shall be 
said to be in its standard position. 

Thus, Fig. 73 represents the three angles 
of Fig. 72 in their standard positions ; the 
angles A OB, AOD, and A OF of Fig. 73 

being equal to the three angles AO x B, C0 2 D, and E0 3 F of Fig. 72 

respectively. 

EXERCISE XXXII 

Place the following angles in their standard positions : 

1. 15°, 225°, 415 3 , 768°. 

2. - 25°, - 275°, - 615°, - 365°. 

3. If two angles differ by an integral multiple of 360° and both 
angles are placed in standard position, how will the terminal sides of 
the two angles be situated with respect to each other ? 

4. If two angles, placed in standard position, have the same terminal 
side, what is the relation between them? 

5. If the sum of two angles is an integral multiple of 360°, how will 
the terminal sides of the two angles be situated with respect to each 
other, both angles being placed in standard position ? 

62. The notion of the trigonometric functions of a general 
angle. Having formulated the notion of a general angle, it be- 
comes necessary to revise our definitions of the trigonometric 
functions, since our original definitions are applicable to acute 
angles only. To be sure, we have already made some prog- 
ress in this direction by defining the functions of an obtuse 
angle. (See Arts. 40 and 42.) But those definitions were 
provisional, and it will be advisable to reopen the whole ques- 
tion, so as to gain a larger and more adequate point of view. 

Our new concept of an angle, as a measure for the amount 
of rotation of a line, practically forces upon us the following 
considerations which automatically suggest the new defi- 
nitions of the trigonometric functions. 



RECTANGULAR COORDINATES 137 

Let us draw a positive acute angle (see the Greek alpha- 
bet on page 132) in its standard position (Fig. 74 a). Let 
us choose a point P anywhere on its terminal side and from 
P drop a perpendicular PM to the initial side. Then, in 
accordance with the definitions of the functions of an acute 
angle, we have 
n , . a MP n OM 

Let us now think of the angle 6 as growing. Nothing 

remarkable happens until 6 reaches 90°. At that moment, 

and as the motion continues, our original definitions cease to 

be applicable, because the 

right triangle POM, of 

which 6 is an interior angle, 

ceases to exist. But we 

may think of PMQ (in 

Fig. 74 a) as a plumb line 

attached to a point P on 

. . . , . , Fig. 74a Fig. 746 

the moving terminal side 

of the angle. If there is no obstacle at 0, this line, re- 
maining always vertical, will pass from the right to the left 
of as 6 grows from an acute into an obtuse angle (see 
Fig. 74 5), and the line-segment OM will change its direc- 
tion. To indicate this change of direction of OM, we rep- 
resent OM by a positive number in Fig. 74 a and by a 
negative number in Fig. 74 b. If we count the distance OP 
(which does not change) as positive, in all positions of the 
moving line, and if we retain equations (1) as definitions 
for sin 6 and cos 6, we see that cos becomes negative when 
6 becomes obtuse. The line-segment PM does not change 
its direction until grows beyond 180°. Therefore the sine 
of an obtuse angle, like that of an acute angle, is positive. 
The sine of an angle, however, becomes negative when the 
angle lies between 180° and 360°. 

63. Rectangular coordinates. All of these things may be 
stated more briefly by the introduction of rectangular coordi- 



138 



FUNCTIONS OF THE GENERAL ANGLE 



M 



Fig. 75 



nates, a notion of utmost importance, not merely in trigonom- 
etry, but in other branches of mathematics. 

Let us draw two lines, unbounded in length and perpen- 
dicular to each other. We shall usually think of one of them 
as horizontal and call it the as-axis, and call the other, which 
is vertical, the t/-axis. The point 0, in which the two axes 
_ intersect, is called the origin of 

coordinates. 

We adopt a unit of length, and 
denote the distances from any point 
P to these two axes by x and y re- 
spectively. In Fig. 75 we have 

NP = 0M= x, MP = 0N= y, 

~ x where the notation is chosen in such 
a way that x is measured on or par- 
allel to the #-axis, and y on or 
parallel to the y-uxis. 
We call x the abscissa and y the ordinate of the point P. 
Both numbers together are called the coordinates of P. 

If we take into account only the magnitudes and not the 
directions of the lines OM, OiV, etc., that is, if x and y are 
regarded as numbers without sign, there will be four points 
which have the same 
coordinates. 

For instance, the points P, 
P', P", P'", in Fig. 76, would 
all correspond to x = 3, y = 2. 

In order to avoid this 
inconvenience, we intro- 
duce the convention that 
the abscissas of all points 
to the right of the ^-axis 
shall be positive, and of 
those to the left negative ; that the ordinates of all 
points above the a>axis shall be positive, and of those below 
negative. 









+ 


y 










p 






+ 


2 






P 










+ 


1 










_ 


3 


-2 


-1 


+i 


+2 


+3 































- 


l 










P" 








2 






P'" 





Fig. 76 



RECTANGULAR COORDINATES 



139 



The coordinates of the four points in Fig. 76 are now different from 

each other. 

The coordinates of P are x = + -5, y = + 2, 

The coordinates of P' are x = — 3, y = 4- 2, 

Tlie coordinates of P" are a: = — 3, y = — 2, 

The coordinates of P'" are x = + -5, // = - 2. 

The positive directions of the x- and y-axes, which have now 
been defined, will hereafter be indicated by a plus sign (as in 
Fig. 76> 

The distance from the origin 0. to any point Pis usually 
denoted by r and is called the radius vector of that point. 
We shall always regard the radius vector as positive, and 
clearly we shall alwaj's have (see Fig. 77), 
r = + V:r 2 + y 2 . 

Let us think of OP (Fig. 77) as rotating around as a 
center. If OP originally coincides with the positive rr-axis, it 
will require a counterclockwise rota- 
tion of 00°, or a clockwise rotation of 
270°, to bring it into coincidence with 
the positive ^-axis. We naturally think 
of the numerically smaller angle first, 
and define the positive sense of rotation 
to be that one which enables us to turn 
the positive .r-axis into the position of 
the positive //-axis by means of a rota- 
tion of only 90°. But this implies that 
the positive direction of rotation is counterclockwise. 
Art. (30.) 



+ 


V 


r^- 




P 
V 







•'' 


J 





Fig. 77 



(Cf. 



EXERCISE XXXIII 

Plot the points whose coordinates are given in Exs. 1 to 5. Find, by 
measurement, to the nearest degree for the angles and to the nearest tenth 
of a unit for the distances, the radius vector of each point and the posi- 
tive angle which it makes with the positive g-axis. Find the same 
results by calculation, making use of three-place tables. 

Li = 4 : '>- y = + b 3. x = - 2.1, y = -J 5.5. 



+ 1: 



5 



+ 1.27. 



4. ./• = - 2.6, y = 
y = -248. 



140 



FUNCTIONS OF THE GENERAL ANGLE 



In the following examples, r denotes the radius vector of a point P, 
and the positive angle which this radius vector makes with the posi- 
tive direction of the z-axis. Plot the points. Find, by measurement to 
the nearest tenth of a unit, the abscissas and ordinates of these points. 
Find the same results by calculation with three-place tables. 

6. r -= 2, $ = 30°. 8. r = 3, = 210°. 

7. r = 5, = 135°. 9. r = 4, 6 = 285°. 

10. r = 2.56, = 310° 20'. 



64. Definition of the trigonometric functions of a general 
angle. We are now in a position to give the definitions of 

the functions of a general angle in 
a compact manner. 

Place the angle 6 in its standard 
position; that is, with its vertex 
on the origin and its initial side 
on the positive x-axis of a system 
of rectangular coordinates. (See 
+ a? Fig. 78, where is a positive acute 
angle.) Pick out a point P, differ- 
ent from the origin, anywhere on 
the terminal side of the angle. Then we adopt the folloiving 
definitions : 




Fig. 78 



The sine of angle 
The cosine of angle 6 = 
The tangent of angle = 
The cotangent of angle = 
The secant of angle 6 = 
The cosecant of angle 6 = 



ordinate of P 
radius vector of P' 

abscissa of P 
radius vector of P 

ordinate of P 
abscissa of P 

abscissa of P 
ordinate of P\ 

radius vector of P 
abscissa of P 

radius vector of P 
ordinate of P 



sine = ^, 
r 



cos u = — . 
r 



tan = 



cot0 = 


~ v 


sec 9 = 


r 

~ QC 


esc 6 = 


r 

= y 



EXCEPTIONAL CASES 141 

As Fig. 78 shows, these definitions reduce to the familiar 
definitions o( Art. 7 if is an acute angle. In the case of 
an obtuse angle, they give results which agree with the defi- 
nitions o( Arts. 40, 42, and 43. But in our present defini- 
tions, is not restricted either in magnitude or sign; it may 
be a positive or negative angle of any magnitude. The 
quantities x and y may be positive, zero, or negative, but r 
is always positive. 

EXERCISE XXXIV 

Construct carefully the following angles and, by measurement, find 
approximate values of the six trigonometric functions, correct to two 
significant places, paying particular attention to their signs. 

1. 25°. 2. 320°. 3. 110. 4. -1:50°. 5. +725°. 6. -10°. 

7. Iu Art. 11, the exact values of the functions of 30°, 45°, and 60° 
were expressed by means of radicals. In a similar way find the values 
of the functions of the following angles: 

120°, 135°, 150% 210°, 225,° 240°, 300°, 315°, 330°. 

8. What are the signs of the trigonometric functions of the follow- 
ing angles: 

150°, 320°, 1000°, - 625°. 

Find, by construction and measurement to the nearest degree, the 
values of the angles for which 

9. sin = - i, cos 6 = + § . 

10. tan 0=1. cos0= L. 

V2 

11. Show that tho vaiues of the trigonometric functions of a general 
angle, as given by the definitions of Art. 64, will not be changed if, 
instead of the point P, any other point P' on the terminal side of the 
angle be chosen. 

65. Discussion of the exceptional cases. Kadi of the 
trigonometric functions is defined in Art. 64 formally^ as a 
quotient of two numbers. This formal definition will have 
a real significance whenever the two numbers actually have 
a quotient. Now we know from Algebra that two numbers, 
D (the dividend) and d (the divisor), always have a unique 



142 FUNCTIONS OF THE GENERAL ANGLE 

quotient q if the divisor is different from zero. That is, 
there exists a number q such that 

(1) J = * 

or what amounts to the same thing, such that 

(2) D = dq, 

whenever d is different from zero. 

Now, let us discuss the case where d (the divisor) is equal 
to zero, while D (the dividend) is not. In this case there 
exists no number q which satisfies equation (2). For this 
equation now becomes 

(3) D = • q, 

and its right member is equal to zero no matter what 
number we substitute for g, while its left member is, b}' 
hypothesis, different from zero. Consequently it involves 
a contradiction to assume that a number has been obtained 
by dividing another by zero, and the operation of dividing by 
zero is therefore excluded from Algebra. 

The formal definitions of Art. 64 involve divisions by x, 
y, and r, and therefore lose their significance in any case in 
which one of these divisors is equal to zero. Now r = OP 
is the radius vector of a point P which may be chosen any- 
where on the terminal side of the angle except at 0. (Com- 
pare the wording of the definitions in Art. 64.) Therefore 
r is never equal to zero. From this fact and the equation 

(4) x 2 + y 2 = r 2 , 

we conclude further that a; and y cannot both be equal to zero 
at the same time. 

It will now be clear that the formal definitions of Art. 64 
fail to provide the symbols 

tan 270°, sec 270°, 
cot 180°, esc 180°, 

But they do define each of the 
six trigonometric functions of all positive angles less than 
360° with the eight exceptions just mentioned. 



(5) 


tan 90°, 
cot 0°, 


sec 90°, 

esc 0°, 


with 


any actual 


meaning. 



EXCEPTIONAL casks 143 

If the angle is greater than 360°, or if it is negative, 
other exceptional eases appear. But their relation to the 
eight exceptional cases (5) is so simple that we may leave 
it to the student to complete this discussion. 

Although the tangent of 90° is not defined, our definitions 
are clearly applicable to the tangent of an angle 6 which 
differs from 90° by the slightest conceivable amount. Let 
us see how the tangent of an angle 6 behaves when ap- 
proaches 90° as a limit. We have, 
in Fig. 79, 

f a V MP 

tan 6 = •' = — — :, 

x OM 




where P may be any point different 

from on the terminal side of the 

angle. For our present purpose it 

will be convenient to select the 

point P in the following manner. Draw a line RS parallel 

to the .r-axis at any convenient distance from OM, and let 

P be the intersection of the terminal side of the angle 6 

with this fixed line. Then, as approaches 90°, MP = y 

will remain constant, while OM=x approaches the limit 

zero. The quotient ^ = tan 6 will therefore become larger 
x 

and larger. Since OM— x may be made as small as we 

please by taking the angle 6 = MOP close enough to 90°, 

while the ordinate MP always remains the same, we see 

that the quotient can be made as large as we please. In 

other words, the angle 6 can be made to differ so little from 

90° that its tangent will become larger than any number 

whatsoever. This is what is meant by the statement that 

tan 6 becomes infinite when 6 approaches 90° as a limit. We 

sometimes express this same statement by writing 

(6) tan 90° = so, 

a symbolic equation which should be interpreted as a short- 
hand account of the situation which has just been described. 
It is not a definition of tan 90°. For x is not a number, and 



144 FUNCTIONS OF THE GENERAL ANGLE 

the symbolic equation (6) is not at all concerned with what 
happens to tan when 6 is equal to 90°. It merely tells us, 
in symbolic form, what happens when 6 approaches 90° as 
a limit ; namely, that tan 6 then increases without bound. 

In the preceding discussion we considered a variable angle 
MOP which approached 90° as a limit. In order to see what 
happened to its tangent, we chose the point P on the terminal 
side of the angle in such a way that its distance y from the 
a>axis remained constant. 

We may obtain the same result in a slightly different way, 
which may, to some students, appear more conclusive. Let 
the angle MOP approach 90° as before. 
Then (Fig. 80), 

tanJfOP = — = &. 
OM x 

Let us, this time, choose the point P on 
the terminal side of the angle in such a 
x way that its distance x from the ?/-axis 
remains unchanged while 6 approaches 90° 
as a limit. It is evident from the figure 
that MP = y will then increase without bound. Therefore, 
we obtain again the result that tan 6 = y/x becomes infinite 
when 6 approaches 90° as a limit. 

So far we have tacitly assumed that 6 is an acute angle 
increasing toward 90° as a limit. What happens when 
starts as an obtuse angle to decrease toward 90° as a limit? 

Since the tangent of any angle between 90° and 180° is 
negative, an argument precisely similar to that just carried 
out shows that the numerical value of tan 6 again grows 
beyond all bounds when 6 approaches 90°, remaining, how- 
ever, always negative. We see therefore that the following 
statements are both true : 

1. When 6 is acute and increases toward 90° as a limit, tan 0, 
remaining always positive, grows numerically beyond bound. 

2. When 6 is obtuse and decreases toward 90° as a limit, 
tan 0, remaining always negative, grows numerically beyond 
bound. 




THE FOUR QUADRANTS 



145 



These two statements are frequently summed up in the 
symbolic formula 
(7) tan 90° = ± ao . 

A precisely similar discussion will show that tan 6 again 
becomes infinite when approaches 270°, that sec 6 becomes 
infinite when approaches 90° or 270°, and that col 6 and 
esc 6 become infinite when 6 approaches either 0° or 180°. 
The functions sin and cos 6 are always finite. 

66. The four quadrants. The x- and ?/-axes divide the 
plane into four portions called quadrants. The quadrant 
bounded by the positive x- and j/-axes is usually called 
the first quadrant. If we start from the first quadrant and 
describe a path around the origin in the counterclockwise 
direction, we traverse in order the 1st, 2d, 3d, and 4th 
quadrants. 

An angle is said to be in the first, second, third, or fourth 
quadrant according to the quadrant in which its terminal 
side falls when the angle is in its standard position, that is, 
with its initial side upon the positive z-axis. 

The cardinal angles 0°, 90°, 180°, 270°, etc., may be regarded 
as belonging to either one of the two quadrants upon whose 
boundaries they lie. 

The following table gives the signs of the trigonometric 
functions of an angle in the various quadrants : 





I 


it 


in 


IV 


Sine 


+ 


+ 


- 


- 


Cosine .... 


+ 


- 


- 


+ 


Tangent . . . 


+ 


- 


+ 





Cotangent . . . 


+ 


- 


+ 


Secant .... 


+ 


- 


- 


+ 


Cosecant . . . 


+ 


+ 


- 


- 



146 



FUNCTIONS OF THE GENERAL ANGLE 



EXERCISE XXXV 

1. Prove each of the following symbolic statements and explain its 
significance in words. 

cot 0° = ± oo , tan 90° = ± oo , cot 180° = ± oo , tan 270° = ± oo , 

csc0 Q =±co, sec 90° =±oo, cscl80° = ±oo, sec 270° = ± oo . 

2. Show that the numerical value of the sine or cosine of an angle can 
never exceed unity. 

3. Show that D/d may have any value whatever if D and d are both 
equal to zero, and hence that the symbol § is wholly indeterminate. Why 
can no one of the trigonometric ratios ever have this form? 

4. Determine the quadrants of the following angles and the signs of 
their trigonometric functions. 

325°, 710°, 1045°, 609°, 412°, - 52°. 

5. In what quadrant is an angle if its sine and cosine are both posi- 
tive? If its sine is positive and its tangent negative? If its secant and 
tangent are both positive ? 

6. If we know that the sine and cosine of an angle have the same 
sign, what can we say about the quadrant of the angle ? 

7. Is there an angle whose tangent is positive and whose cotangent is 
negative ? 

8. If we are told that the tangent and cotangent of an angle are both 
positive, does this enable us to determine the quadrant of the angle ? 



67. General character of the trigonometric functions. Their 
periodicity. We are now in a position to understand how the 

functions change with the angle. 
For the purposes of this discussion 
it will be convenient to think of the 
radius vector r as constant. This 
means that the point P, which ac- 
cording to our definition must be 
selected on the terminal side of the 
angle, describes the circumference of 
FlG - 81 a circle as $ changes from 0° to 360°. 

(See Fig. 81.) It is easy to verify the following statements 
by reference to the figure : 



+ 


y 




\ ° 




x Ml 



PERIODICITY OF FUNCTIONS 



14' 



As increases from 0° bo 90°, sin increases from <> to 1. 
A- 6 increases from 90° to 180°, sin decreases from 1 to 0. 
As tf increases from ISO to :27<)°, sin decreases from to — 1. 
As 6 increases from -70° to 300°, sin 6 increases from —1 to 0. 

It is also evident that the function sin repeats its values 
inexactly the same order if P moves around the circumference 
a second, third. ••• nth time. The same thing is true of the 
other trigonometric functions, a very important fact which 
may he expressed as follows: 

Each of the si.v trigonometric functions is periodic and its 
period is equal to 3(30°. That is. each of these functions repeats 
its values at i)itervals of 360°, so that 

sin {0 + n • 3o0°) = sin 0, cos (6 + n • 360°) = cos 0, etc., 

where n is any positive or negative integer or zero. 

The behavior of each of the six functions in the neighbor- 
hood of each of the four cardinal angles may be recapitulated 
for convenience of reference in the following table : 





S I N K 


COSINI 


Tangbott 


COTANGKNT 


Ski ANT 


COSBOANT 


0° 





+ 1 





T x 


+ 1 


Tx 




+ 1 





± x 





± x 


+ 1 


180 J 





- 1 





Tx 


- 1 


± x 


870 


- 1 





± x 





Tx 


- 1 



The student should use this table to describe in words the 
variation of each of the six functions as changes from 0° 
S60°. 

EXERCISE XXXVI 
I)i-cu<s the variation of the following functions as varies from <> to 

1. cos0. 6. Bin .2 ft 11. sin (- 0). 

2. tan 6. 7. -J .in 6. 12. cos (- 0). 

3. cotft 8. sin 3ft 13. Bin | 0. 

4. sec 0. 9. Bin 1 6. 14. sin ■ 6. 

5. CSC ft 10. tan 10. 15. sin (0 + 25°). 



148 FUNCTIONS OF THE GENERAL ANGLE 

68. Relations between the trigonometric functions of a 
general angle. The relations which we found in Art. 9, 
between the functions of an acute angle, still hold without 
alteration for an angle of any magnitude. In fact the equa- 
tion between the abscissa, ordinate, and radius vector of a 
point P, that is, x 2 _j_ 7 ,2 _ r ^ 

is true, no matter in what quadrant the point P may be 
situated. 

This is due to the fact that only the squares of x, y, and r occur in this 
relation. 



If we divide both members of the above equation by r 2 , 
we find (A 2 >(y\ 2 =1 

But, by the definitions of Art. 64, we have, in all quadrants, 



- = cos 6, ^ = sin 0, 
r r 



so that the preceding equation 


becomes 


(1) sin 2 8 + cos 2 8=1. 


Since we have, by definition, 




sin = ^, 
r 


esc u = — , 

y 


COS0 = -, 

r 


sec a = -, 

X 


tan 6 = &, 

X 

we find at once 


COt0 = -, 

y 



(2) sin6 esc 8 = 1, cos 8 sec 8=1, tan 8 cot 8=1. 

We have also , a _y _ y/r _ si n 
x x/r cos#' 
which, combined with (2), gives the further relations 

^o\ u. o sin 8 .a cos 8 

(3) tan 8 = — — , cot 8 = -. 

cos 8 sin 8 

If we divide both members of (1) by cos 2 6 and make use 
of (2) and (3), we find 

(4) 1 + tan 2 8 = sec 2 8. 



TRIGONOMETRIC tDENTlTIES 149 

In a similar fashion, if we divide both members of (1) by 

sin 2 0, we see thai 

(5) 1 +eot-e = csc-0. 

When the angle is acute, all of its functions are positive. 
Consequently if one oi' its functions is given, all of the others 
may be found without ambiguity by means of the above rela- 
tions. (Cf. Art. 9 and Exercise VI, Exs. 7-12.) 

But if we do not know in what quadrant an angle lies and 
are given the value of merely one of its functions, the angle 
itself and its other functions are not determined uniquely. 

If we are told, for instance, that sin = £, equation (1) only tells us 

that cos-$= l - (\y = |; 

so that cos $ = ± \ V:3, 

where either sign may be taken. In fact there are two angles between 

i» ami 360° whose sines are equal to £; namely, 30° and 150°. We may 

distinguish between them by stating whether the cosine is positive or 

negative. 

EXERCISE XXXVII 

Find the other functions of the angle 6 as determined by each of the 
following conditions : 

1. sin = — 2 and is in the third quadrant. 

2. -in = — \ and is in the fourth quadrant. 

3. tan $ = + 2 and is in the third quadrant. 

4. cot $ = — 3 and sin is positive. 

5. s( j c = + 2 and tan is negative. 

6. Find the values of the other functions if sin0 = a. Are all pos- 
sible values of a admissible? State a reason for your answer. 

7. If tan#= in. find the values of the other functions. 

8. If -t j c 6 = k, find the values of the other functions. Are all values 
of k admissible in this problem? State a reason for your answer. 

69. Trigonometric identities which involve functions of a 
single angle. By means of the relations of the preceding 
article, an expression which involves the trigonometric func- 
tions of an angle 6 may Ik; written in a great many different 
forms. It is often important to be abb' to recognize that 
two trigonometric expressions, although different in form, 



150 FUNCTIONS OF THE GENERAL ANGLE 

are really identical. This may frequently be done by inspec- 
tion. In more complicated cases it is advisable to express 
each of the two quantities, whose identity we wish to estab- 
lish, in terms of some one of the six functions (the sine, for 
example). It will then become evident as a mere matter of 
algebra whether or not the two quantities are really 
identical. 

EXERCISE XXXVIII 

1. Show that sec 6 — tan sin = cos#, for all values of 6 for which 
tan 6 and sec $ are denned. (See p. 142.) 

Solution. We have, for all values of 0, for which tan and sec 6 are 
denned, 

a j. a • a 1 sin . n 1 — sin 2 cos 2 a 

sec — tan 0sm 6 = -sm = — — — = cos 0, 

cos 6 cos cos 6 cos 6 

which proves the truth of the original assertion. 

2. Prove that tan A + cot A = sec A esc A is an identity.* 
Solution. Denote the quantity on the left member by L and that on 

the right member by R. Then 

L = ton A + cot A = ^A + ^A = ^ A + ™* 2 A = ± 

cos A sin A sin A cos A 

R = sec A esc A = — — = -. 

cos A sin A sin A cos A 

Therefore L = R; that is, 

tan A + cot A = sec A esc A. Q. e. d. 

Prove that the following statements are identities : 

3. cos 6 tan = sin 0. 7. cos 2 A - sin 2 A =2 cos 2 A - 1. 

4. sin <fiCot<j> = cos c£. 8 - (esc 2 6-1) sin 2 6 = cos 2 9. 

5. sin 2 6 + sin 2 6 tan 2 = tan 2 0. 9. 1 + tan 2 = ^— • 

1 — sin 2 

6. cos 2 ^l-sin 2 J=l-2sin 2 /l. i . C os 4 — sin 4 = 2 cos 2 - 1. 

11. (sin + cos $Y + (sin - cos 6>) 2 = 2. 

12. secflcotfl-cscfltanfl^ csc fl seca 

cos — sin 6 

13 coBflcotfl-sinfltanfl =1 + 8in co8ft 
esc — sec 



14. aM ?HL^ = sec $ _ tan 0. if is an acute angle. 

y 1 + sin 6 

*In other words, show that the left member is equal to the right member for 
all values of A for which the functions tau.4,cot^4, sec A, esc J. have been 
defined ; that is, for all values of A except A = 0°, 90°, 180°, 360°, etc. 



CHAPTER X 

GRAPHIC REPRESENTATIONS OF THE TRIGONO- 
METRIC FUNCTIONS 

70. Line representation of the trigonometric functions. The 

trigonometric functions were defined as ratios or abstract 
numbers, not as lines. (See Art. 7.) This does not, how- 
ever, preclude the possibility of representing them as lines. 
An abstract or concrete quantity of any kind may be repre- 
sented as a line-segment, by choosing arbitrarily a certain line- 
segment to represent a unit of the same kind. 

Thus we may represent as lines the populations of the various states 
of the Union, taking a line-segment one inch long to represent a popula- 
tion of 1.000.000. The populations of New York and Illinois will then 
be represented by line-segments 0.11 and 5.64 inches in length respec- 
tively. Thus, although a population obviously is not a line-segment, it 
may be represented by a line-segment. In the same way we may repre- 
sent the values of the trigonometric functions by lines, although they are 
not lines, but abstract numbers. 

The following is a convenient method for obtaining a 
representation of the values of the trigonometric functions 
as line-segments. 

We construct a circle with the origin of coordinates as 
center. An angle whose vertex is at the center of this circle 
will subtend an arc whose numerical measure, in degrees, 
minutes, and seconds, is equal to that of the angle. We 
may therefore speak indifferently either about the functions 
of the angle or of the functions of tin? arc. The point in 
which the initial side of the angle meets the circle is called 
the origin of tin' arc. The point in which the terminal side 
of the anjHe meets the circle is called the terminus. <>r the 
end of the arc. 

161 



152 GRAPHIC REPRESENTATIONS OF FUNCTIONS 



If an angle is placed in its standard position, the origin of 
the subtended arc will be at A, the point in which the positive 
#-axis meets the circle. We shall call this point the primary 
origin of arcs. The point B, in which the positive ?/-axis 
intersects the circle is called the secondary origin of arcs. 

Let us choose any convenient unit of length, say an inch, 
and let us agree to measure all distances in terms of this 
unit. We then construct the circle, the so-called unit circle, 
whose center is at the origin of coordinates and whose radius 
is equal to the unit of length. 

In Fig. 82, let AOQ — 6 be any angle in its standard posi- 
+?/ tion, and AP the arc which it sub- 

b tends on the unit circle. Then 




a) 



+x 



sin 6 
cos 6 



y 



MP, 
OM, 



since the distance r = OP is equal 
to the unit of length, so that r = 1. 
Now x = OM and y = MP are the 
coordinates of P, the terminus of the arc AP. Conse- 
quently we may express our result as follows : 

If any angle 6 is placed in its standard position, the value of 
its sine is equal, in magnitude and sign, to the ordinate of the 
terminus of the arc zvhich the angle subtends on the unit circle. 
The value of its cosine is equal to the abscissa of the terminus 
of this arc. 

In order to find a line representation for tan 6 and cot 0, 
we draw tangents to the unit circle at A and B and denote 
by T and T' the points in which the terminal side of the 
angle intersects these two tangents. (See Fig. 83.) 

Then we find 

AT^AT 
OA 1 



tan = — — = 



AT, 



cot 6 = tan BOT' = 



BT ! BT' 



OB 



1 



= BT f , 



LINE REPRESENTATION OF FUNCTIONS 



lf>3 



since the radius of the circle 

<)A = OB=l. 

If $ is an obtuse angle, OP will have to be prolonged 

backward in order to intersect the tangent at A. Moreover 
the point of intersection 
T will then be below .4. 
Now the tangents BT' and 
AT are parallel to the x- 
and //-axes respectively. 
Let us agree to give signs 
to the line-segments meas- 
ured on these two tangents 
as though they were ab- 
scissas or ordinates of a 
point. That is, let AT be 

positive or negative according as T is above or below A, 
and let BT' be positive or negative according as T' is to 
the right or left of B. This convention is indicated in 
Fig. 83 by the two + signs at the ends of the two tan- 
gents. The student may now verify that the equations 




Fig. 83 



(2) 



tan 6= AT, cot 6=BT, 



which we have obtained from Fig. 83 in the case of an acute 
angle, will give correct results in magnitude and sign, no 
matter in what quadrant the angle 6 may happen to fall. 

We may formulate our results as follows : 

If any angle 6 is placed in its standard position, the value of 
its tangent is equal, in magnitude and sign, to the ordinate of 
the point in which tin- terminal side of the angle, prolong^] 
backward vfm cessary, intersects the tangoit t<> the unit circle at 
tin- primary origin of arcs. 

The cotangent of tin- angle is equal, in magnitude and sign, 
to tin- abscissa of tin- point i,, which tin- terminal side of the 
angle, prolan vard if necessary, intersects tin- tangent to 

the unit circle at tin- secondary origin <>f <<,- .<. 



154 GRAPHIC REPRESENTATIONS OF FUNCTIONS 



Referring once more to Fig. 83, we have 



(3) 



a OT OT nm 



esc 6 = sec BOT' 



OT' 



= 0T>. 



OT' = 
OB 1 

These line representations for the secant and cosecant will 
hold, in magnitude and sign, not merely for acute angles, but 
for angles in any quadrant, if we agree to make the following 
conventions in regard to sign. OT shall be positive if T is 
on the same side of as P, i.e. if T is on the terminal side 
of the angle 6. OT shall be negative if ^is on the terminal 
side of the angle 6 prolonged backward. OT' shall be 
positive or negative according as T' falls on the terminal 
side of the angle 6 or on the terminal side prolonged back- 
ward. 

We leave it to the student as an exercise to verif}^ these 
statements in detail and to formulate the contents of equations 
(3) in words. 

Figures 84 to 87 illustrate the line representation of the 




GRAPHS OF FUNCTIONS L55 



trigonometric functions for an angle 
quadrants. In each of these figures 


in each one o 


MP = sin 6, AT = tun 0, 
0M= cos 6, BT = cot 0, 


0T= seed, 

or' = csc o. 




These line representations of tan and sec suffice to explain why the 

Dames tangent and secant were chosen for these functions. The word 
sine is not capable of such a simple explanation and has a long and 
complicated history. 

The Greeks did not use the six functions which we 
have introduced. In place of the sine of an angle they 
made use of the chord PQ. subtended by the angle POQ, 
on a circle of known radius. (See Fig. 88.) If the 
circle has a unit radius. Fig. 88 shows that this chord 
PQ is equal to twice QR, or Fig. 88 

PQ= 2 sin j POQ. 

Thus the chord, used by the Greeks, is essentially twice the sine of 
half the angle. 

Aryahiiata. a famous Hindoo mathematician (born 476 a.d.), was 
apparently the first to introduce the sine of the angle in place of the 
chord, and he, quite naturally, spoke of it as the half-chord or jyd-ardhd, 
where jya is the Sanskrit for chord or bowstring and ardha for one half. 
For the sake of brevity the adjective ardha was soon omitted and the 
sine was called simply jya. 

The Arabs, who far more than any other people cultivated the sciences 
during the Middle Ages, took over this word from the Hindoos, but 
changed its spelling to jiba, so as to make the spelling accord with the 
pronunciation in the sense of their own language. 

But in written Arabic the consonants only are represented by definite 
characters, the vowels being merely indicated by dots which are fre- 
quently omitted altogether. As a consequence of this practice, the Hin- 
doo word jiba was soon corrupted into jaih, a genuine Arabic word 
meaning bosom, heart, or pocket according to the context. 

In the twelfth century, when the Arabic texts were translated into 
Latin, the word jatb was translated literally by the Latin word sinus 
meaning bosom. 

Thu<. a foreign word was first converted by the Arabs into a word of 
their own language having a Bimilar sound but an entirely different 
meaning, and later this Arabic word was translated literally into Latin. 
Of course the derivation of the English word sine from sinus is obvious. 

71. Graphs of functions, a number of whose numerical 
values are given. There i> a second way <>f representing 



156 GRAPHIC REPRESENTATIONS OF FUNCTIONS 



graphically the values of the trigonometric functions, which 
is even more important than that which has just been dis- 
cussed. For it gives us, in a still more vivid fashion, a 
picture of all the most essential properties of these functions ; 
and it has the further advantage of being applicable, not 
merely to the trigonometric functions, but to all of the other 
functions which naturally arise in pure and applied mathe- 
matics. In order to lead the student to appreciate fully the 
power of this new method, we shall first illustrate it by 
a number of examples taken from fields other than trig- 
onometry. 



x =■ Date 


y = Population 


1790 


3,929,214 


1800 


5,308,483 


1810 


7,239,881 


1820 


9,638,453 


1830 


12,860,692 


1840 


17,063,353 


1850 


23,191,876 


1860 


31,443,321 


1870 


38,558,371 


1880 . 


50,155,783 


1890 


62,947,714 


1900 


75,994,575 


1910 


91,972,266 




1790 1800 1810 1820 1830 1810 1850 1860 1870 1880 .1890 1900 1910 

Fig. 89 



The population of the United States is determined every ten years 
by a national census. The table in the margin gives the results of these 
censuses. It is customary to represent the contents of this table graph- 
ically by laying off the dates horizontally {i.e. as abscissas), and erect- 
ing for each of these dates a vertical line (ordinate), which shall give 
by its length in terms of an appropriately chosen unit the population at 
that time. Figure 89 gives such a graphic representation of the facts 
contained in this population table and presents these facts in a more 
easily intelligible form than the table itself. Moreover if we join the 
endpoints of the ordinates by a smooth curve (the population curve), 



GRAPHS OF FUNCTIONS 



157 



we may draw some fairly reliable conclusions as to the state of the 
population in the years 1805, L815, etc., in which no census was taken. 

If we plot the population curves of two or more countries upon the same 
sheet, a great many interesting matters may be brought out by comparison. 

Clearly we may adopt such a graphic method, whenever 
we have a table giving a relation between two variables ; 
that is. a table which shows that to certain numerical values 
of a first quantity x there correspond certain numerical values 
of a second quantity y. 

EXERCISE XXXIX 

1. The following table gives the population of the cities of New York, 
Chicago, and Philadelphia for the years named: 



New York . 
Chicago . . 
Philadelphia 



1850 



I860 



L8T0 



515,547 805,651942,292 
28.2C)!) 109,206298,9 

340.o4o.385.520(374.022 



L880 



1,206,200 
503.208 
847,170 



1,515,301 
1.000.850 
1,046.004 



1900 



3,437,202 
1,698,572 
1,293,697 



lino 



4,766,883 
2,185,283 
1,540.008 



Make a graph illustrating this information,* and from the graph find 
the probable population of each of these cities in 190S. 

2. Make a population table and a population curve for the city and 
in which you live. 

3. Let the student provide himself with a railroad time-table, giving 
the names of the various stations, their distances from the starting point, 
and the times at which a certain train leaves these stations. Draw a 
distance-time diagram for one or several trains, plotting the times as 
abscissas and the distances as ordi nates. 

4. On April 3. 1012, the following temperatures were observed in 
Chicago : 



3 A.M. 


32 


11 A.M. 


38 


7 P.M. 


3 1 


1 A.M. 


32 


12 M. 


39 


S P.M. 


33 


5 A.M. 


32 


1 P.M. 


39 


P.M. 


31 


A.M. 


31 


2 P.M. 


38 


10 P.M. 


:;i 


7 A.M. 


32 


3 P.M. 


37 


11 P.M. 


33 


- A.M. 


33 


1 P.M. 


36 


12 P.M. 


33 


'.» A.M. 


M 


.". P.M. 


35 


1 A.M. 


32 


1<I A.M. 


36 


6 P.M. 


.15 


2 A.M. 


31 


I; iphically. 











* In all snen work, involving plotting of enrvee, it i- advisable t<> use cross- 



158 GRAPHIC REPRESENTATIONS OF FUNCTIONS 

72. Graphs of simple algebraic functions. The relation 
between the variables x and y, instead of being given by 
a table as in the previous examples, may be given by an 
equation. We may then use the equation for the purpose 
of constructing a table, and then draw a graph as before. 



Example 1. Find the graph of y = 2 x — 5. 



X 


_ o 


- 1 





+ 1 


+ 2 


+ 3 


+ 4 


+ 5 



Solution. 

y 



-7 
— 5 
-3 
-1 

+ 1 
+ 3 
+ 5 



If we substitute x = in the given equation, we find 
y =— 5; for x = 1, we find y = — 3 ; etc. AVe construct 
in this way the table printed in the margin. If we plot the 
points x— — 2, y~— 9; x = — 1, y = — 7; etc., obtained 
in this way, we find the points marked in Fig. 90 with a 
little cross. All of these points are found to be on a straight 
line. 

This observation makes it seem likely that all of the 
points whose coordinates satisfy the equation y=2x — 5, 
not merely those which we happened to compute, are on this 
same straight line. It is not difficult to prove that this is 
so, but the proof will not be given here. 




+ 


/ 


, 


+10 / 


T 


49 ^ 


1 


t8 / 


\ 


-K / 


\ 


1-0 / 


\ 


+•} / 


A 


+± r 


\ - 


+3 / 


\ - 


+2 / 


• \ 


+■1/ 



Fig. 90 



-4-3-2-1 +L+2 +3^-4 

Fig. 91 



±x 



Example 2. Find the graph of y = x 2 . 

Solution. As before, we construct the table in the margin by comput- 
ing the values of y which correspond to the values x = — 3, — 2, — 1, 0, 



X 


-3 


-2 


- 1 





+ 1 


- 2 


+ 3 



GRAPHS OF TRIGONOMETRIC FUNCTIONS 159 

// + 1, +2, +3. We then plot the points obtained in this 
way, the points marked with a little cross in Fig. 91, and 

unite them by a smooth curve. 
4 
l The principle involved in these examples, that to 

every equation between two variables x and y there 

corresponds a curve and vice versa, is at the founda- 

tion of Analytic Geometry, which is one of the most 

important developments of modern mathematics. 

The great merit of having introduced this idea into 

mathematics is due to Descartes (1596-1650) and Fekmat 

(1601-1665). 

EXERCISE XL 

Find the graphs of the following equations : 

1. y = x. 8. y = 2 x 3 - 3 x + 1. 

2 II = ~ X. 9. y = X 3. 

3. y = - 2 x + 1. 10. y = * s* - L 

4. y = 2*». 

5. // = — ./-. 3 

6. y = ./-' 2 — 5. -. 

12. y = ±-l. 

7. y = ./'- — 5 x. x 

73. Graphs of the trigonometric functions. We now pro- 
ceed to apply this method to the relation 

y = sin x. 

We choose an arbitrary line-segment on the z-axis to repre- 
sent one degree and another arbitrary line-segment on the 
y-axis to represent the unit value of the sine, that is, the 
abstract number 1. If we are using millimeter paper, or a 
metric scale, it will be convenient to make a distance of one 
millimeter on the z-axis stand for one degree, and to measure 
the ordinates in terms of a unit 10 centimeters or 100 mil- 
limeters long. As this is a rather large scale it will probably 
be necessary to paste several sheets together in order to be 
able to construct the whole our, 



160 GRAPHIC REPRESENTATIONS OF FUNCTIONS 



From the table of natural sines we obtain the table in 
the margin, in which the values of the angle x as well as 
the corresponding values of sin x are expressed in milli- 
metres in accordance with the adopted scale, which makes 
1 mm. on the rr-axis stand for 1°, and 100 mm. on the y- 
axis stand for the unit value of the sine. This table 
enables us to plot ten points of our curve represent- 
ing ten values of the function sin x in the first quad- 
rant. (See Fig. 92, which is a reduced copy of such a 
curve.) 

The definition of the sine of a general angle 

(Art. 64) and the line representation of the 

sine in the unit circle (Art. 70), both show very clearly 

that two angles like 80° and 100°, or 70° and 110°, which 



X 


y = sin x 








10' 


17 


20 


34 


30 


50 


40 


64 


50 


76 


60 


87 


70 


94 


80 


98 


90 


100 




+x 



Fig. 92. — The Sine Curve 



differ by the same amount from 90° but in opposite direc- 
tions have the same sine. Consequently that portion AB 
of our curve, which represents the values of sin x for angles 
in the second quadrant, will be a symmetric counterpart of 
the first portion OA, which corresponds to angles in the 
first quadrant. (See Fig. 92.) 

The unit circle also makes it evident that the sines of two 
angles which differ by 180° are numerically equal but 
opposite in sign. Consequently that portion BOD of our 
curve, which corresponds to angles in the third and fourth 
quadrants and all of whose ordinates are negative, may be 
obtained easily from the known part OAB. The parts OAB 



GRAPHS OF TRIGONOMETRIC FUNCTIONS 



llil 



and BCD of the curve are in fact congruent, but are situated 
on opposite sides of the .r-axis. 

We have already noted that the sine function repeats its 
values at intervals of 360°. It is a periodic function with a 
period of 3C>0° (Art. 67). This manifests itself in the graph 




; + X 



Fig. 93. — The Cosine Curve 

by the fact that the picture in each of the intervals from 
360° to 720°, etc., from - 360° to 0°, etc., is an exact copy 
of that piece of the curve which lies between 0° and 360°. 

The curve obtained in this way from the sine function is 
called the sine curve. 

The cosine curve, which is the graph of the function 

y = cos £, 

is of the same general character as the sine curve. Its form 
is given in Fig. 93, and may be -obtained by applying to the 
cosine function an argument exactly similar to that which 
has just been carried out for the sine. 

We may apply the same method to the function 

y = tan x. 

However, the graph obtained in this way differs very essen- 
tially from the sine and cosine curves. 

In fact we know that when x approaches 90 c from below, 
that is. if x assumes a succession of values like 

89°, 89°.l». 89°.99, 89°.999, etc., 



162 GRAPHIC REPRESENTATIONS OF FUNCTIONS 

the tangent of x, remaining always positive, will grow numeri- 
cally beyond all bound. If on the other hand x approaches 
90° from above, through a sequence of values like 

91°, 90°.l, 90°.01, 90°.001, etc., 

the tangent of x, remaining always negative, again grows 
numerically beyond all bound. (Cf. Art. 65.} We see, 
therefore, that the difference between the values of 

tan (90° + K) and tan (90° - K) 

grows larger and larger as the angles 

90° + h and 90° - h 

themselves come closer and closer together. 

We express this by saying that the function tan x is discon- 
tinuous for x = 90°. The corresponding property of the 
graph is an interruption or break in the otherwise continuous 
curve. There are no such breaks in the sine or cosine curves. 
We can think of a material point (say the point of a lead 
pencil) as actually describing a sine curve without interrup- 
tion. If we were to attempt to do the same thing for the 
tangent curve, we should have to interrupt the path of the 
point at x = 90°, at x = 270°, etc. 

We meet here the important distinction between continuous 
and discontinuous functions, the precise formulation of which 
must be left to a later point in the student's career. 

The tangent is, of course, a periodic function and repeats 
its values at intervals of 360°. But we may now observe 
that, unlike the sine or cosine, it repeats its values after the 
shorter interval of 180°. To recapitulate: the tangent is a 
periodic function of period 180°, and is discontinuous for 
x = 90° and for all values of x which differ from 90° by integral 
multiples of 180°. 

Figure 94 shows the form of the tangent curve. 

EXERCISE XLI 

1. Plot the curves y = 2 sin x, y = 3 sin x, y = 4 sin x. 

2. Plot the curves y = sin 2 x, y = sin 3 x, y = sin 4 x. 



RADIAN MEASURE 103 

3. How are the corves o( Exs. 1 and 2 related to the curve 

H = sin x? 

4. Show that the curve // = cot ./■ is discontinuous for x = 0°, 180°, 
etc., ami has the form indicated in Fig. 95. 




Fig. 94. — The Tangent Curve 



Fig. Dj. — The Cotangent Curve 



5. Show that the curves y = sec x and y = esc x have the forms indi- 
cated in Figs. 96 and 97. 






Vu, 96.— The Secant Curve 



Fig. 97. —The Cosecant Curve 



74. The natural unit of circular measurement. Definition 
of a radian. In constructing the graphs of the trigonometric 
functions, the student may have observed that the units of 
measurement on the x- and y-axes were both chosen arbi- 
trarily, and might have been selected in infinitely many 
different ways, thus altering materially the appearance of 
the resulting curve. 

This mutual independence of the two scales, on the two 
coordinate axes, is a natural consequence of the facl that 



164 GRAPHIC REPRESENTATIONS OF FUNCTIONS 

the two quantities x and y were regarded as different in 
kind. One of them was regarded as an angle measured in 
degrees, and the other as an abstract number. Having 
chosen a certain horizontal line-segment as representative 
of the unit of angles (1°), it was still admissible to choose 
arbitrarily another (vertical) line-segment to represent the 
unit of abstract numbers. Whenever x and y represent two 
quantities of different kind, the x- and ^-scales are, in the 
nature of things, independent of each other. 

Even if x and y are quantities of the same kind, it is often 
more convenient to choose the lengths of the units different 
on the two scales. If this were not done, the resulting 
curve might fail utterly to serve the purposes for which it 
was intended. 

Thus, when we draw a profile map of an extensive country (showing 
the elevations of various points in a certain vertical cross section), the 
vertical scale must be chosen much larger than the horizontal scale. 
Otherwise the differences of elevation, as depicted on the map, would 
become so small as to be unnoticeable. 

Nevertheless it will usually be desirable to choose the hori- 
zontal and vertical units equal to each other whenever x and 
y may be regarded as quantities of the same kind, provided 
that the resulting curve does not thereby lose its usefulness 
as it would in the example just quoted. 

Now the considerations of Art. 70 show that, from a cer- 
tain point of view, the quantities x and y which occur in 
such an equation as .. _ • „, 

may be regarded as quantities of the same kind. 

In fact, we observed in Art. 70 that we might think of 
the number x as the measure of the arc AP (Fig. 82) in- 
stead of as the measure of the corresponding angle AOP. 
We saw further that certain line-segments could be con- 
structed whose lengths, in terms of the radius of the circle 
as unit, were equal to the values of sin x, cos x, etc. If then 
we measure the length of the arc x in terms of the radius 
of the circle as unit, instead of in degrees, we shall have the 



RADIAN MEASURE 105 

arc and its trigonometric functions expressed in terms of the 

same unit. 

TIi is unit of arc measure, an arc of a circle whose length is 
equal to the radius of the circle, is called a radian. 

One advantage gained by measuring arcs in radians is 
this: the arc ami its trigonometric functions will then be 
expressed in terms of the same unit. 

If r is the radius of a circle, the length of its circumfer- 
ence is equal to 2 irr. Therefore a circumference may be 
said to contain 2 it radians. Since it also contains 360 de- 
grees, we have 

(1) 2 it radians = 360°, 

whence 

1 r 360° 180° 
1 radian = = . 

2 IT IT 

Since it = 3.14159265, we find, to seven decimal places, 

(2) 1 radian = 57°. 2957795. 
On the other hand we find from (1) 

(3) 10 = lio radians ' 

or 

(4) 1° = 0.0174533 radian. 

These equations make it easy to find the number of de- 
grees in an angle or arc when its measure is given in radians, 
or vice versa. 

Clearly it follows, from the definition of a radian, that the 
length of the arc which an angle of one radian intercepts on 
the circumference of a circle of radius r is itself equal to r. 
Then, an angle of half a radian at the center will intercept 
an arc on the circumference whose length is equal to \ r. 

In general : <ni angle of 6 radians 'ft the center of a circle of 
is r intercepts an <ir<- % upon the circumference whose length 
is 

(5) s = rO. 



166 GRAPHIC REPRESENTATIONS OF FUNCTIONS 

In the simplicity of this formula lies a second great ad- 
vantage of measuring angles in radians rather than in de- 
grees. 

EXERCISE XLII • 

Convert into radians the following angles: 

1. 90°. 3. 30°. 5. +693° 20'. 

2. 270°. 4. + 25° 15'. 6. - 1030° 0'. 

Convert into degrees the following angles which are given in radians, 
state their quadrants and the signs of their, trigonometric functions : 



7. 7r . 

2 


9. 3L. 

16 


11. 0.7691. 


8. *. 

4 


10. 3.14159. 


12. 5.3214. 



13. Prove that the area of a sector of a circle of radius r is equal to 
\ r 2 if 0, the angle at the center, is measured in radians. 

14. Prove that a segment of a circle of radius r, whose arc is equal 
to 6 radians, has the area 

ir 2 (0-sin0). 

15. Compute the area of a circular segment of radius 11 feet, if its 
arc is equal to 52°. 

16. How will the formulae of Exs. 13 and 14 be modified, if the angle 
6 is expressed in degrees ? 

17. A cord is stretched around two wheels, a large one of radius r 
and a smaller one of radius r' feet, the distance between the centers of 
the wheels being d feet. If the cord is not crossed and if 6 is the angle 
of inclination, expressed in radians, of the free part of the cord to the 
line of centers of the wheels, show that 



[ d cos6+(l + 0)r + (?-ey], 



d ' 
where I is the entire length of the cord. 

18. If the cord in Ex. 17 is crossed, show that its length I may be 
found by means of the formulae 

sin = L±lL, i = 2 pcos 6 + (? + fl) (r + r')l- 

19. Find {he length of a belt which is to be stretched around two 
wheels 3 and 2 feet in diameter respectively, if the distance between the 
centers of the two wheels is 5 feet : (a) if the belt is crossed, (b) if it is 
not crossed. 



FUNCTIONS OF SYMMETRICAL AXGLES 



1(17 



20. Draw the graphs of t ho trigonometric functions sin r, cos r, tan r, 

cot .r. it ./■ is expressed in radians, using the same unit of Length for- 
distances and y distances. 



75. Relations between the functions of two symmetrical 
angles. Let us consider two angles, like 90°— 6 and 90° + 0, 

which differ from one of the cardinal angles by the same 
amount but in opposite directions. If we place two such 
angles in the standard position, their terminal sides will be 
symmetrically situated with respect to one of the two 
coordinate axes, so that this axis will bisect the angle be- 
tween them. As a consequence of this fact the trigo- 
nometric functions of the two angles are related to each 
other in a very simple fashion. 

In order to obtain these relations we shall consider each 
of the four cardinal angles separately, making use of the 
line representation of the functions 
given in Art. 70. 

Let ns begin with the cardinal 
angle 0° or radians. Figure 98 
represents the unit circle and the 
two angles 

6 = Z.A0P and - 6 = Z AOP' , 

each in its standard position. The 

arcs AP and AP' subtended by fig. 98 

these angles on the unit circle are 

symmetrical with respect to A. Therefore the ordinates 

of P and P' (the termini of these arcs) are numerically 

equal and opposite in sign, while their abscissas are equal in 

magnitude and sign. 

But the ordinate and abscissa of the terminus of the arc 
AP are respectively equal to the sine and cosine of 6. 
The ordinate and abscissa of P' are respectively equal to the 
sine and cosine of — 6. (See Art. 70.) 

Consequently we have 
(1; sin( - 0)= -sin (9. cos ( - #) = cos 0. 




168 



GRAPHIC REPRESENTATIONS OF FUNCTIONS 



Consider next the case of two angles 90° — and 90° + 
symmetric with respect to the cardinal angle 90°. In this 

case (see Fig. 99) P and P' have 
the same ordinate, while their ab- 
scissas are numerically equal but 
opposite in sign. Consequently 

sin (90° -0) = sin (90° + 0), 




(2«) 



cos(9O°-0)=-cos(9O° + 0), 



Fig. 99 



IT 



or, if the angles are measured in 
radians, 



(2 5) sin(|--0 



if . 



*)cos(f-*)= 



M^ + 



By a precisely similar argument, the details of which we 
leave to the student, we find 



(3 a) 
or, 
(3 5) 



sin (180° - 0) = - sin (180° + 0), 
cos (180° - 0) = cos (180° + 0) ; 

sin (tt — 0) = — sin (w + 0), 

C0S(7T — 0) = COS (7T + 0), 



according as the angle is measured in degrees or radians, 
and also 



(4 a) 
or, 

(4 5) 



sin (270° - 0) = sin (270° + 0), 
cos (270° - 0) = - cos(270° + (9) ; 



V^-0 



f3* 

cos — 
V2 










Of course we shall also have 

sin (360° - 6) = - sin (360° + 0), 
cos (360° - 0) = cos (360° + 0). 



RELATIONS AMONG THE FUNCTIONS 169 

But these equations are really repetitions of (1) if we 
remember that, on account of the periodicity of the sine 
and cosine, 

sin (800° + 0) = sin 0, sin (360° - 0) = sin ( - 0), 
cos (360° -f 0)= cos 0, cos(360° - 0)=cos(- 0). 

The relations, which correspond to (1), (2), (8), (4) for the 
remaining trigonometric functions, may easily be obtained 
by expressing the tangent, cotangent, secant, and cosecant in 

terms of the sine and cosine. (See Art. 08.) Thus, for 
instance, 

, /dn sin( — 0) — sin 6 D 

tan ( — #) = * -£ = — = — tan 6. 

J eos(-#) cos# 

The student should actually work out the sixteen equations 
obtainable in this way and combine them in tabular form 
with the eight equations (1) to (4). 

The student should also observe that, although- we have 
constructed the figures for the case when 6 is a positive acute 
nngle, our proof of formulae (1), (2), (3), (4) remains valid 
word for word, if 6 is a positive or negative angle of any 
magnitude. A good way to convince one's self of this fact is 
to think of the angle 6 as variable and to follow out mentally 
the changes which would take place in such a figure as 
Fig. 98 or Fig. 99 when the angle 6 increases or decreases. 
The fact that equations (1) to (4) are universally valid will 
thus be rendered intuitive. 

76. Relations between functions of two angles whose sum 
or difference is a right angle. If 6 is an acute angle, we 
know, from Art. 10, that 

(1) sin (90° -0)= cos 0, cos (90° -0) = sin 0. 

If we combine these equations with (2 a) of Art. 75, we find 
further 

(2) sin (90° + 0) = cos 0, cos (90° + &) = - sin 0. 

We wish to show that the four equations (1) and (2) are 
true, not merely when is an acute angle, but when 6 is a 



170 GRAPHIC REPRESENTATIONS OF FUNCTIONS 

positive or negative angle of any magnitude. This may be 
done by the method of mathematical induction. 

We begin by proving the following theorem. If equations 
(1) and (2) are true for a certain angle 0, they are also true 
for the angle 0' = 90° + 0. 

Proof. By hypothesis, equations (1) and (2) are true 
for the angle 0. Therefore we have 

sin & = sin (90° + 0) = cos 0, cos 0' = cos (90° + 0) = - sin 0. 
But 

sin (90° - 00= s in [90° -(90° + 0)] = sin(- 0)= - sin 0, 
and 

cos (90° - 0') = cos [90° - (90° + 0)] = cos ( - 0) = cos 0, 

(Art. 75, equations (1)), 
which proves that 

(3) sin (90° - 00 = cos 0', cos (90° - 00 = sin 0', 

since both members of the first equation are equal to — sin 0, 
and both members of the second are equal to cos 0. 

Since equations (2 a) of Art. 75 are true for all angles, we 
now find 

sin (90° + 00 = sin (90° - 0') = cos 0', 
cos (90° + 0') = - cos (90° - 00 = - sin 0' . 

Since equations (3) and (4) are the same as (1) and (2), 
with 6' in p]ace of 0, we have actually proved our theorem; 
namely, if equations (1) and (2) are true for the angle 0, 
they are also true for the angle 6' = 90° + 0. 

We know that equations (1) and (2) are true for all posi- 
tive acute angles. As a consequence of the theorem just 
proved, they are successively seen to be true for all positive 
angles in the second, third, or fourth quadrant, and conse- 
quently for all positive angles whatever. 

But they are also true for all negative angles. For let 
be a negative angle. Let n ■ 360°, where n is a positive in- 
teger, be the lowest integral multiple of 360° which makes 

0' = + n • 360° 



THE QUADRANTAL FORMULA 



171 



a positive angle. Then, on account of the periodic character 

of the sine and cosine, we shall have 



,.-,, 



and similarly 



sin 0' = sin (0 + n • 360°)= sin 0, 
cos = cos( 4- n • 360°) = cos 0, 



(6) 



sin(9O o -^)=sin(9O o -0), cos(9O°-0') = cos(9O°-0), 



sin(9O°+0')=sin(9O o +0), cos(9O° + 0' ) = cos(9O° + 0). 

Since ^' is a positive angle, we have 

Bin (90° - 0') = cos 0', cos (90° - 0')= sin 0', 
sin (90° + 0') = cos 0', cos (90° + 0')= - sin 0' . 

If in these equations we substitute the values (5) and (6), 
we rind 

sin (90° - 0)= cos 0, cos (90° - 0) = sin 0, 
sin(90° + 0)= cos 0, cos (90° + 0)= - sin 0. 

Therefore equations ( 1) and (2) are true for positive and nega- 
tive angles of any magnitude. 

The formulae for tan (90° + 0), sec (90° + 0), etc., may be 
found by expressing these functions of 90° 4- in terms of 
sin (90° + 0) and cos (90° + 0) and making use of (2); for 
instance, we find 

tan (90- + 0) = si " (° ° + *> = -°™g. = - cot 0. 
' cos (90° + 0) -sin0 

77. The quadrantal formulae. If we unite equations (1) of 
Art. ~t> with the corresponding formulie for the remaining 
four functions, we obtain the following system of equations: 

sin (90° - 0) = cos 0, cos (90° - 0) = sin 0, 

(1) | tan (90° - 0) = cot 0, cot (90° - 0)= tan 0. 

sec (90° - ) = esc 0. esc (90° - 0) = sec 0. 

In the same way we find, from equations (2) of Art. 76, 
the system: 

sin (90° + 0) = cos 0, cos (90° + 0) = - sin 0, 

(2; { tan (90° + 0) = - cot 0, cot(9O° + 0)=- tan 0, 

90° + 6) = - esc 0. esc (90° + 0) = - sec 0. 



172 GRAPHIC REPRESENTATIONS OF FUNCTIONS 

Since these equations are true for angles of any magnitude 
and not merely for acute angles, we conclude from (1) and 
(2) that 



sin (180° - 0) = sin (90° + 90° - 0) = cos (90° - 0) = sin 0, 
cos(18O o -0) = cos(9O o + 9O o -0)=-sin(9O°-0) = -cos 0, 

etc., giving rise to the further system of equations : 

f sin (180° - 0) = sin 0, cos (180° - 0) = - cos 0, 

(3) tan (180° - 0) = - tan 0, cot (180° - 0) = - cot 0, 
I sec (180° - 0) = - sec 0, esc (180° - 0) = esc 0. 

In similar fashion we find 



sin (180° + 0) = sin(90° + 90° + 0) = cos (90° + 0) = - sin 0, 
cos (180° + 0) = cos (90° + 90° + 0) = - sin (90° + 0) 

= —cos 0, etc., 
so that we obtain 

r sin (180°+ 0)= - sin 0, cos (180° + 0)= - cos 0, 

(4) tan (180° -f 0) = tan 0, cot (180° + 0) = cot 0, 

I sec (180° + 0) = - sec 0, esc (180° + 6) = - esc 0. 

Again we have 

sin (270° - 0) = sin (90° + 180° - 0) = cos (180° - 0) 

= — cos 0, etc., 
whence 

f sin (270° - 0) = - cos 0, cos (270° - 0) = - sin 0, 

(5) tan (270° - 0) = cot 0, cot (270° - 0) = tan 0, 

I sec (270° - 0) = - esc 0, esc (270° - 0) = - sec 0, 

and similarly 

sin (270° + 0) = - cos 0, cos (270° + 0) = sin 0, 

(6) tan (270° + 0) = - cot 0, cot (270° + 0) = - tan 0, 
. sec (270° + 0) = esc 0, esc (270° + 0) = - sec 0. 

Finally we find the equations 

f sin (360° - 0) = - sin 0, cos (360° - 0) = cos 0, 

(7) I tan (360° - 0) = - tan 0, cot (360° - 0) = - cot 0, 
[ sec (360° - 0) = sec 0, esc (360° - 0) = - esc 0, 



THE QUADRANTAL FORMULA 173 

and the system 

sin(360 + 0) = sin0, cos (360° + &) = cos 0, 
(8) ■ tan(36O°+0) = taD 0, cot (360° + 0)=cot 6\ 

. S ec (360° + 0) = sec 0, CSC ( 360° + 0) = esc 0, 

which latter equations merely express the periodic character 
of the trigonometric functions. 

Since, on account of the periodicity of the functions, we 
ha ye 

sin (360° - 6 ) = sin (- + 360°) = sin ( - 6), etc., 

we may also write, in place of (7), 

r sin( — 0)= — sin 0, cos(— 0)= cos (9, 
tan(-0)=-tan0, cot(- 0)=- cot 0, 

[ sec (— 0) = sec 0, esc ( — 0) = — esc 0. 

The 48 formulae ^1) to (8), the so-called quadrantal formu- 
lae, have a very important practical application. They serve 
the purpose of finding the values of the functions of angles not 
situated in the first quadrant. 

For instance, if we wish to find the .sine and cosine of 310°, we may 
use equations (6), which give 

sin 31 if = sin (270° + 40°) = - cos 40°, 
008 310° = cos (270° + 40°) = sin 40°. 

The numerical values of cos 40° and sin 40° may, of course, be taken 
from the tables. 

On account of the practical application just mentioned, it 
is important to be able to remember the 48 quadrantal 
formulae. This is not at all difficult if we impress upon our 
minds some of their peculiarities. 

We observe in the first place that all of the angles which 
appear in the left members of these equations are of the form 

a multiple of 90° ± 6. that is, a cardinal angle ± 6, 

while the angle which appears in the right member is al- 
ways simply 6. 

Let us speak of those cardinal angles, like 90° and 270°, 
which are odd multiples of < . ,(>a as odd cardinal angles, while 



174 GRAPHIC REPRESENTATIONS OF FUNCTIONS 

the even cardinal angles, like 0° and 180°, are even multiples 
of 90°. 

If now we look through our list of quadrantal formulas, we 
observe that they are all included in one of the two forms : 



(10) 



function of (even cardinal angle ± 0) 

= ± same function of 0, 

function of (odd cardinal angle ± #) 

= ± corresponding co-function of 6, 



it being understood, as in Art. 10, that the six functions are 
arranged in three pairs, sine and cosine, tangent and cotan- 
gent, secant and cosecant, each member of each pair being 
regarded as the co-function of the other. 

We have observed, then, that the same function occurs in 
both members of a quadrantal formula whenever the correspond- 
ing cardinal angle is even. If the cardinal angle is odd, the 
function which appears in the right member is the co-function 
of that one which appears on the left. 

It remains to describe a method for remembering which of 
the two signs, + or — , should be used in any one of these 
formulae. We may determine this sign by thinking of the 
particular case when 6 is a positive acute angle. The quad- 
rant of the angle in the left member of the equation will then 
be evident by inspection, and therefore also the sign of the 
left member. The ambiguous sign ±, on the right member, 
must then be chosen in such a manner as to make the two 
members of the equation agree in sign. 

An example will make this clear. We wish to find the formula for 
sin (270° + 0). Since 270° is an odd cardinal angle, we have in the first 

P lace sin (270° + 0) = ± cos 6. 

To determine the sign on the right member, we think of the case when 
6 is a positive acute angle. Then 270°+ 6 is in the fourth quadrant, and 
sin (270° + 0) is negative, while cos 0, being the cosine of a positive acute 
angle, is positive. Therefore we must choose the — sign, so that 

sin (270° + 0) = - cos 0, 

since choice of the + sign would lead to the absurdity of equating a 
positive to a negative number. 



THE SINK AND COSINE CURVES L75 

EXERCISE XLIII 
Express the following as functions o\ positive acute angles : 

1. cos(-75°). 3. tan 517°. 5. cot 175°. 

2. sm325°. 4. csc(-412°). 6. sec 1562°. 

7. Express the functions of Exs. 1-ti as functions of positive acute 

angles less than 1.". . 

8. Compute the value of the expression 2shl(80+lO o ) for the 
values of = 25 . •"»<» . 75 . 100°. 

Find the values of the following expressions : 

9. cos <i<> cos 120° - sin 60° sin 120°. 
10. sin :;n cos 300° + cos 30° sin 300 . 



11. tanl^tan 1177 



ir+-m-m- 



Ij 

12. cos 315° sin 11° - tan 293° see 25 . 
Simplify the following expressions : 

13. ,/- + //-+ 2db cos (180 - x). 

14. (a - I) tan (90° + .1) + (a + b) cot (- A). 

15. „ sin(^ -#) + /, cos (tt-<9). 

16. tan + tan (tt - $). 

Stat*- for what values of each of the following expressions is positive, 
and for what values of 6 it is negative : 

17. sin 6 - cos 0. 19. Bin 2 - cos 2 6. 

18. sin + co> 6. 20. tan - cot 0. 

21. Find the formulae for the functions of 6 — - in terms of the 
functions of 0. the angle $ being measured in radians. 

22. Find formulae for the functions of 6-tt in terms of the functions 
of ft 

78. Properties of the sine and cosine curves. The prop- 
erties of the sine and cosine which were discussed in Arts. 
75, 7''.. 77 manifest themselves very clearly if we make use 
of the graphs of these functions as obtained in Art. 73. We 
shall slightly modify these graphs, however, by thinking of 



176 GRAPHIC REPRESENTATIONS OF FUNCTIONS 

the angle x as being measured in radians (see Art. 74) rather 
than in degrees, and by choosing the same unit of length to 

+?/ 















[' 




P" 








/b' 


c 


\b' 


A' M' 


0/ 






A 




B M'"C 


A) 


/-2ir 


— Sir 
2 


Vt 


rr 1 / 





M 


■km" 
2 


7rV 


Bit 
2 ^ 


A* 








^ F' 












P' 







Fig. 100. — The Siue Curve. Natural Scale 

represent one radian on the a>axis as that which represents 
the abstract number 1 on the ^/-axis. 

Figure 100 represents the sine curve 

y = sin x 
constructed in accordance with this choice of units. 

Let us consider two points of this curve which are at the 
same distance from the ?/-axis but on opposite sides, such as 
P and P' . The ordinates of these points are obviously 
numerically equal but opposite in sign. Denote the abscissa 
of P by z ; then, that of P f will be — g, and we find that 
(l) s sin(— z) = — sing. 

If we draw a line parallel to the #-axis through the point 

A for which x = ^, this line clearly divides the curve into 

two symmetrical portions. Consequently two points, such 
as P and P", at the same distance from this line but on 
opposite sides of it, will have equal ordinates. That is, 
sin OM= sin OM' '. 
But if we denote MA by 2, we have 



so that 
(2). 



OM=l-z, 



I \~ z 



OM" = ^ + z, 

2 



l| + s 



THE SINE AND COSINE CURVES 177 

Let the curve be divided into two portions by a line parallel 
to OF through the point B whose abscissa is equal to it. 
Points of the curve at equal distances from this line but on 
opposite sides of it, such as P" and P'", have ordinates 
numerically equal but opposite in sign. Therefore 

(3), sin(7r — z ) = — sin(7r + z). 

In similar fashion we find 

(4). sin (^ - z^j = sin (^ + z). 

(5), sin(2?r- z) = — sin (2 *-'+*)- 

The points M and M" were equidistant from .4. Conse- 
quently the distances OM and M" B are equal. Since the 
ordinates MP and M" P rf are equal, we shall have 

sin OM" =sin OM. 

If we put 0M= z, we shall have M" B = z, and therefore 

OM " = 7T - 2, 

so that the preceding equation becomes 

(6) 8 sin(7r — z) = sin z. 

By similar considerations in connection with the cosine 
curve, we find a system of relations which correspond com- 
pletely to the above equations (l) s to (6) a . They are 

(l) c cos ( — z) = cos z, 

(2) c cos (| - z J = - cos ( | + z 

(3) c cos (it — z) = cos (tt + z), 

(4) c cos f-y - 2 J = - cos (-^ + z 

(5) c cos(2w — g) = cos(2w + s), 

(i>) c COS QlT — Z) = — COS 2. 

The truth of all these equations which are thus suggested 
by the curves has been established, in a slightly different 
notation, in Arts. 75-77, 



178 GRAPHIC REPRESENTATIONS OF FUNCTIONS 

We can hardly fail to notice the striking similarity between 
the sine and cosine curves. In order to put into evidence 
the relations between them, we construct Fig. 101 which 
contains them both. 



+y 




+<*? 



Fig. 101 



This figure suggests that, if we displace the cosine curve 



7T 



toward the right through a distance of — units, it will coin- 

z 

cide with the sine curve. If this is true, a point P of the 
cosine curve whose coordinates are 



00 



OM= z, MP = cos z, 



and which by our displacement will be brought into coin- 
cidence with a point P' whose coordinates are 



CJ) 



0M> = z + |, M'P> = MP, 



should, in its new position, be a point on the sine curve. 
Therefore the coordinates, OM' and M'P', of this point P' 
should satisfy the relation y = sin x which is satisfied by the 
coordinates of all points of the sine curve. This gives 



(<0 



MP = sin ( z + | ), 



and, on combination with (a) and (5); 



(7). 



sin ( z + Z ) = M'P' = MP = cos z. 



TIIK SINK AND COSINE CURVES 179 

Thus, equation (7), must be true if the geometric relation 
lift ween the sine and cosine curves suggested by Fig. 101 is 
actually based on fact. But this equation coincides with 
formula (2) of Art. 76, except for the notation, so that its 
validity is no Longer open to question. 

Consequently, the sine and cosine curve differ only in posi- 
tion and may be brought into coincidence by a displacement of 

units parallel to the x-axis. 

EXERCISE XLIV 
1. What geometric property of Fig. 101 corresponds to the relations 
sin f — — z J = cos z, cos f — — z J = sin z ? 



2. How are the relations 
sin 
to be obtained from Fig. 101? 



^ + *) = -COS*,COs(^ + *) = 



3. Plot the tangent and cotangent curves and discuss these graphs in 
a fashion analogous (so far as possible) to the discussion of Art. 78. 



CHAPTER XI 



RELATIONS BETWEEN THE FUNCTIONS OF MORE THAN 

ONE ANGLE 



79. The addition theorems for sine and cosine. In Art. 77 
we expressed sin ( 6 + — ), cos ( 6 -f ^ ), etc., in terms of sin 



and cos 6. 



The angle 6 was an angle of any magnitude, but 

the angle added to it was always an integral multiple- of — 

radians or 90°. The question now arises whether it is pos- 
sible to find similar formulas for sin (ot + /3) and cos (a + /3), 
where both a and /3 are angles of any magnitude. 

Let us assume, to begin with, that a and /3 are both positive 
acute angles whose sum a + ft is also acute. We place the 
acute angle a in its standard position 
xOA. (See Fig. 102.) We then 
place the angle {3 with its initial side 
upon OA (the terminal side of the 
angle a), so as to make Z A OB equal 
to /3. Then 

ZxOB= a + /3, 
and moreover this angle is in its stand- 
ard position. Therefore if we take 
any point P, different from 0, on its terminal side OB and 
drop a perpendicular PM from P to the a>axis, we shall have 




(i) 



sin (a + /3) = 



MP r y ct\ 0M 

— , cos(«+/3) = ^. 

Let us drop perpendiculars PQ, QN, and QR from P to 
OA, from Q to the #-axis, and from Q to MP. Then we have 



(2) 



sin(. + ^)=^ = M±^ == Z2 



OP 



OP 



+ 



i£P 



OP OP 



180 



ADDITION THEOREMS FOR SINE AM) COSINE 181 

Now \ is a ratio of sides of two different right triangles, 

namely, ONQ and OPQ. But these triangles have the side 

OQ in common, and this common side may be used to trans- 

XO 
form —j- into a product of two ratios, each of which contains 

two sides of the same right triangle and is therefore a trigo- 
nometric function of the acute angles of this triangle. In 
fact we find 

NQ NQ OQ Q 

( 3) — ^ =^~ - -rr^- = sin a cos p. 

OP OQ OP 

In the same way we rind for the second term of (2) 

RP^RP PQ 
OP PQ ' OP' 



But 



— ^- = sin /3, and — — = cos RPQ = cos a, 



OP ' PQ 

since the sides of the angle RPQ are respectively perpen- 
dicular to those of a. Consequently we find 

( 4 ) — — = cos a sin /3. 

If (8) and (4) be substituted in (2), we obtain the impor- 
tant formula 

sin (« -f/8)= sin a cos /3 4- cos a sin /3. 

Referring once more to Fig. 102, we have 

*M{m + tl).- op - op - Qp Qp . 

If we again transform each of these ratios into a product 
of two others, we find 

ON ON OQ Q 

-—— = • — ^- = cos a cos p, 

op oq op 

BQ RQ PQ . . p 

~~ = — ^ • — - = sin a sin p, 
OP PQ OP 



182 



FUNCTIONS OF MORE THAN ONE ANGLE 



so that 

(7) cos (ot -f- /3) = cos a cos j3 — sin a sin /3. 

It is easy to see that equations (5) and (7) remain valid if 
a and ft are acute angles, even if their sum is greater than a 

right angle. In that case we shall 
have the situation represented in 
Fig. 103. If we make precisely the 
same constructions as before, the 
proof of the formula for sin (a + /3) 
will remain applicable word for word. 
But since a -+- (3 is now in the sec- 
ond quadrant, its cosine is negative ; 
that is, 

MO 




cos (a + /3) = — 



OP 



where by MO we mean merely the positive number express- 
ing the length of the line-segment MO, so that —MO is a 
negative number. 

Now the figure shows that 

MO = MN- ON= BQ- ON, 



so that 



cos (« + £) = — 



MO ON- RQ 



ON 
OP 



RQ 
OP 



OP OP 

From this point on, the proof proceeds exactly as in the pre- 
vious case, beginning from equation (6). 

Thus, we have proved that equations (5) and (7) are cer- 
tainly true if a and ft are positive acute angles, even if their 
sum is greater than 90°. 

We may now show that these formula are true for two 
angles in any quadrant. In order to do this, we first prove 
the following theorem. If the formulce 

(8) sin (a -h /3) = sin a cos /3 + cos a sin /?, 

cos (a + /3) = cos a cos (3 — sin a sin /3, 

are true for two angles a and /3, they remain true if either of 
these angles be increased by 90°. 



ADDITION THEOREMS FOR SINK AND COSINE 183 

PROOF. Let us assume that equations (S) are true for a 
certain pair of angles «. and ft. Tut 

«' = 90 c +«, 
so that 

vi + ft = 90° + a + £. 

We shall then have (Art. 77, equations (2)), 

sin (a* +f$) = sin (90° + a + 0) = cos (a + /3) 
( 9) = cos a cos /? — sin « sin /3, 

cos (at + £) = cos (90° -f- a + £) = - sin (a + /3) 

= — sin a cos ft — cos a sin /3, 

and also 

sin a! = sin (90° + a) = cos a, cos «' = cos (90° + a) = — sin a, 

whence 

sin « = — cos «', cos a = sin a' . 

If we substitute these values in (9), we find 

sin («' + ft) = sin a' cos ft+ cos «' sin ft, 
cos («' + £) = cos a' cos /5 — sin a' sin /3. 

But these formulae are of the same form as equations (8), 
with a' = a + 90° in place of a. The same process would 
show that equations (8) would still be satisfied if we replaced 
ft by ft' = 90° + ft. Consequently our theorem is proved. 

We know already that equations (8) are true if a and ft 
are any two positive acute angles. On account of the 
theorem just proved they will still be true if a is in the 
second quadrant and ft in the first, and hence also if both a 
and ft are in the second quadrant, and hence if one of these 
angles is in the third quadrant, etc. Therefore equations 
| 8 ) are true for all positive angles. 

But they are also true if either or both angles are negative. 
For instance, let a be a negative angle while ft is positive. 
By adding to a a sufficient number n of complete positive 
revolutions, we shall obtain 

a' = a + n- 360°, 



184 FUNCTIONS OF MORE THAN ONE ANGLE 

a positive angle. But then 

sin a' = sin a, cos a 1 = cos a, 

sin (V + /3) = sin (a + /3), cos (a' + /3) = cos (a + /3), 

so that 

sin (a + /3) = sin (a' + /3) = sin a' cos ft + cos a' sin ft 

= sin a cos /3 + cos a sin /3, 

and similarly for cos (a+ /3). 

If /3 is negative, we may proceed in the same manner. 
Consequently we obtain the following important theorem: 

If a and /3 are two positive or negative angles of any magni- 
tude, the sine and cosine of their sum are always given by the 
formulae, 

sin (a + p) = sin a cos p + cos a sin p, 
(8) 

cos (a + p) = cos a cos p — sin a sin p. 

The method of mathematical induction employed for proving 
the general validity of these formulae is of fundamental 
importance in all parts of mathematics. We have already 
made use of it in Art. 76. 

Equations (8) are usually known as the addition theorems 
for sine and cosine. 

EXERCISE XLV 

1. Compute the sine and cosine of 75°. 

Solution. We have 75° = 30° + 45°, sin 30° = §, cos 30° = |VB, sin 45° 

= cos 45° = Therefore 

V2 

sin 75° = sin 30° cos 45° + cos 30° sin 45° = ? 
cos 75° = cos 30° cos 45° - sin 30° sin 45° = ? 

Compute the sines and cosines of the following angles : 

2. 120°. 4. .210°. 6. 195°. 

3. 150°. 5. 105°. 7. 165°. 

8. Find formulae for sin (45° + 6) and cos (45° + 0). 

9. Find formulae for sin (30° + 0) and cos (30° + 6). 



THEOREMS FOR TANGENT AND COTANGENT 185 

10. Find formulae for sin (60S + 6) and eos(60° + 6). 

11. Show that those oi the quadrants] formula? of Art. 77, which in- 
volve the sine and cosine, are special eases of the addition theorems for 
the sine and cosine. 

Prove thai the following equations are true for all values of the angles 
which appear in them. That is, prove that they are identities. 

12. sin (a + /3) sin (it — /?) = sin a a - sin- (3 = cos 2 /J — cos 2 a. 

13. cos (u + (3) cos(a - (3) = cos- a — sin- (3 = cos 2 (3 — sin- a 

= l (cos 2 « + cos 2 j3) - I (sin 2 « + sin 2 /?). 

14. Bin (60° + a)- sin (60° - a)= sin a. 
Simplify the following expressions: 

15. sin (1 + n) cos (1 - n) + cos (1 + n) 6 sin (1 - n) 0. 

16. cos (1 -f n) cos (1 -n) $ - sin (1 + n) sin (1-/0 0. 

80. The addition theorems for tangent and cotangent. The 
addition formulae for the tangent and cotangent may be 
obtained as consequences of those for the sine and cosine. 
We have 

, , a \ sin (a 4- ft) sin a cos ft -f- cos a sin ft 

tan (« + /S) = — ^ — — ^ = !-~- — • 

cos (a -h ft) cos a cos p — sin a sin /8 

The fraction in the right member of this equation may be 

expressed in terms of tan a and tan ft by dividing both 

numerator and denominator by cos a cos ft. We obtain in 

this way 

sin a cos ft cos « sin ft 

, ~ N cos a cos ft cos a cos /3 

tan (a + ft) = : : — 3 , 

-. _ sin a sm ft 

cos a cos ft 

or, finally, 

/i\ / , o tan cl + tan B 

(1) tan (a + p) = - — ^ 

1 — tan a tan p 

By a similar process we find 

cot a cot yS — 1 



(2) cot(« + /3) = 



cot a -f cot ft 



186 FUNCTIONS OF MORE THAN ONE ANGLE 

EXERCISE XLVI 

Find the tangents and cotangents of the following angles : 

1. 75°. 3. 210°. 

2. 105°. 4. 150°. 

Prove that the following equations are identities: 
1 + tan 6 



1 - tan 6 
cot - 1 



5. tan (45° + 0) = 

6. cot (45° + 6) = 

v } cot e + 1 

7. Find formulae for tan (30° + 0) and cot (30° + 0). 

8. Find formulae for tan (60° + 0) and cot (60° + 0). 

81. The subtraction formulae. The formulae for sin (a — /3), 
cos (a — /3), etc., may be obtained from those for siu (a + /3), 
cos (a + yS), etc. For, since the addition formulae are known 
to be true for all positive and negative values of a and /3, we 
may write 

sin(a-£) =sin [a + (-£)] 

= sin a cos ( — /3) + cos a sin (— /3), 

cos (a — /S) = cos [« 4- ( — £)] 

= cos a cos ( — /3) — sin a sin ( — /S) . 

But, on account of equations (9), Art. 77, we have 
sin ( — ft) = — sin j3, cos ( — £) = cos /3, 
so that we find 

sin (a — p) = sin a cos p — cos a sin p, 

(i) 

cos (a — p) = cos a cos p + sin a sin p. 

In the same way we find 

/fn , s a\ tana— tan B 

(2) tan (a — P) = £-, 

v J r 1 -f-tanatanp 

and 

, ON , , on cot a cot /8 + 1 

(3) cot (a— /3) = £ . 

v } v y cot /3- cot a 

Of course equations (2) and (3) may also be obtained 
from (1) by division. 



PRODUCTS AND SUMS OV FUNCTIONS 187 

EXERCISE XLVII 

1. Let the student devise a geometric proof lor equations (1), in the 
case when (c and (3 are positive acute angles, a being the larger of the t\\ «». 

Note. If Fig. L02 he described in words, but if the angle there 
denoted by j3 be instead regarded and constructed as a negative angle, 
the same description will apply to both figures and the same method of 
transforming the ratios which was used in Art. 79 will be effective in the 
present example. 

82. Formulae for converting products of trigonometric func- 
tions into sums, and vice versa. Before the invention of 
Logarithms the ealculations of products and quotients was a 
very laborious process. In "those days, then, it was considered 
a great simplification if a formula whose numerical evaluation 
required multiplication could be transformed into another 
requiring only addition or subtraction.* The addition and 
subtraction formulas will enable us to accomplish this for a 
product of two sines, of two cosines, or of a sine and cosine. 

In fact, we have the two equations 

sin (« + £) = sin a cos ft + cos « sin ft, 
sin (« — ft) = sin a cos ft — cos a sin ft, 

from which we derive, by addition and subtraction, 

1 . sin ( a + ft) -f- sin (a — ft) = 2 sin a cos ft, 

sin (a + ft) — sin (a — ft) — '2 cos a sin ft. 

In the same way. from 

cos (a + ft) = eos a cos ft — sin « sin ft, 
cos (a — ft) = cos a cos /3 + sin a sin /3, 

we tind 

,« cos (a -f /3) + cos (« — /3) = 2 cos a cos /3, 

cos (a + ft) — cos (« — /3) = — 2 sin « sin /3. 

From these last equations we find, by transposition and 

division by -. 

,qn COS u cos ft = I [cos (a — ft) -\- COS (a -f- /3)], 

sin a sin /3 = .1 [cos (a — ft) — cos (a + /3)], 

* This was the so-called prosthapharetic method. 



188 FUNCTIONS OF MORE THAN ONE ANGLE 

whereas, from the first equation of (1), we find 

(4) sin a cos /3 = \ [sin (ot — /3) -f sin (a + /3)] . 

The result obtained from the second equation of (1) is 
the same as (4) except for an interchange of the letters a 
and /3. 

Equations (3) and (4) are important in that they enable 
us to transform a product of two sines, two cosines, or of a sine 
and a cosine into a sum or difference. 

For many purposes this is very important even nowadays,* 
although not for the purposes of numerical calculation. In 
fact, at the present time, for numerical -work we prefer 
formulae which involve multiplication to those involving 
addition, because the former process is more easily performed 
by logarithms. 

The above formulae, slightly modified, may also be used 
for the purpose of converting sums and differences of sines 
and cosines into products. Let us put 

a + 13 = A, a-/3=B, 

so that a = $(A + B), /3 = ±(A-B). 

Then, equations (1) and (2) yield the following four for- 
mulae : 

sin A + sin B = 2- sin |- (A + B) cos J (A — B), 
sin A - sin B = 2 sin ±(A-B) cos J (A + B), 
cos A + cos B = 2 cos 1 (A - B) cos i (A + B), 
cos A - cos B = - 2 sin l(J. - B) sin %(A + B). 

We have seen the first two of these equations before, hav- 
ing derived them directly from a figure (in Art. 49) for the 
purpose of proving the law of tangents. Our proof, at that 
time, did not permit us to affirm that the formulae were true 
for all angles A and B. That such is the case, however, has 
now been made evident, since the present proof was obtained 
without placing any restrictions on the values of the angles 
A and B. 

* For instance, in harmonic analysis (see Arts. Ill and 112) and in the integral 
calculus. 



(5) 



V2 



FUNCTIONS OF DOUBLE ANGLES 189 

EXERCISE XLVIII 
Prove the following equations: 

1. sin •'} + sin V 6 = 2 sin 2 6 cos 0. 

2. sin ^ + x\ + sin f 7 ^ - z) = 2 sin £ cos 

sin 6 <i + sin 1 (c 

3. ! = tan 5 a. 

cos 6 a — cos 4 a 

a - R 
tan 1± 

sin «c - sin /? _ '2___ 

sin a + sin £? tan a + JS 

5. cos a — cos :> a = 2 sin « sin 2 a. 

6. sin 3 (C + cos « = sin 3 « + sin (90° - a) = ? 

7. By generalizing the process observed in Ex. 6, derive a formula 
for sin a + cos /?. 

8. Derive a formula for sin a — cos (3. 

Reduce the following products to sums or differences : 

9. Bin 4 a cos 2 a. 12. cos 2 6 cos 8 6. 

10. ain 6 9 sin 4 0. 13. sin 5 « cos 3 a. 

11. cos 2 sin 4 0. 14. sin- 6 cos 6. 

15. Making use of the formulae (5) of Art. 82, show how to derive 
the law of tangents from the law of sines. 

83. Functions of double angles. If we put /3 = a in equa- 
tiona (S) of Art. 70, we find 

I 1 ) sin 2 a = 2 sin a cos a, 

and 

( 2) cos 2 a = cos' a — sin' a. 

On account of the relation 

Bin 3 a + cos 2 a = 1, 

the latter equation may also be written in either of the fol- 
lowing two forms : 

(3) cos 2 a = 1 — 2 sin 2 a, 
or 

c-i) 



190 FUNCTIONS OF MORE THAN ONE ANGLE 

If we put ft = a in equations (1) and (2) of Art. 80, we 
find 

2 tan a 



(5) tan 2a = 

(6) cot 2 a = 



1 — tan 3 a 
cot 2 a — 1 



2 cot a 

which equations may, of course, also be derived from (1) and 
(2) by division. 

84. Functions of half angles. In Art. 83 we regard the 
functions of a as known, and we learn how to compute the 
functions of 2 a. We shall now invert the problem by 
regarding as known the functions of 2 a, the problem being 
to calculate the functions of a, the half angle. To put the 
character of the problem more clearly into evidence, we shall 

P Ut 2a=e. «=J0, 

which merely amounts to thinking of any angle as a double 
angle; namely, as double its half. 

With this change of notation, equations (3) and (4) of 

Art. 83 become 

(1) cos = 1 - 2 sin 2 :!, cos = 2 cos 2 ^ - 1. 


If we solve the first of these equations for 2 sin 2 „ and the 

q 
second for 2 cos 2 -, we find 

2 

(2) 2sin 2 ^ = l-cos0, 2 cos 2 ^ = 1 + cos 0, 
whence 

. ON . 8 , ^ /i - cos e e /i + cos e 

(3) sin-=±^ , cos- = ±V 

The ambiguous signs on the right members are determined 

q 
by the quadrant of the angle — If 6 is a positive angle not 

6 ^ 

greater than 180°, - is in the first quadrant and the + sign 

A 

must be chosen in both of the equations (3). 



FUNCTIONS OF HALF ANGLES 191 

From (3) we find, by division, 



,x . e /i -cose . e , % \ + 

4) tan- = ±\ /— -, cot- = ±V' 

2 ^l + cosO 2 M - 



the appropriate sign being again determined by the quadrant 
of the angle .! 0. 

According to (1), Art. 83, we have 

(5) 2 sin - cos - = sin 6. 

Let us divide each member of the first equation of (2) by 
the corresponding member of (5). We find 

fKS . l-cos0 

(6) tan - = . • 

2 sm 6 

Bv a similar process we obtain, from the second equation 

of (2), 

.-. .6 l + cos0 

(0 cot-= \ . 

2 sin 6 

These last two formuhe might also have been derived from 
(4). But the proof we have given is preferable since it 
avoids the necessity of discussing the ambiguous sign, a dis- 
cussion which would be necessary if we had followed the 
other method. 

6 

The equations for sin -, cos -, etc., are of very great impor- 
tance, because they may be used for the purpose of computing a 
table of trigonometric functions. In fact, we have already 
shown how to calculate the values of the functions of 0°, 30°, 
45°, 60°, and 90° (Art. 11). By means of the addition and 
subtraction formuhe (Arts. 79, 81) we are therefore in a posi- 
tion to find also the functions of 15° and 75°. We can now 
calculate the values of the functions of one half of 15° or 
7°.5, of one half of 7°.5, or 3°.75, etc. By continuing this 
process of bisection and combining the results by means of 
the addition theorem, we may obviously compute the values 
of the functions for a set of angles between 0° and 90° as 



192 FUNCTIONS OF MORE THAN ONE ANGLE 

close together as we please. By interpolation we may then 
find the functions of 1°, 2°, 3°, etc. 

The method, of which we have just given an outline, is essentially the 
same as that employed by Ptolemy (second century a.d.).* Ptolemy, 
however, also made use of the inscribed pentagon (cf. Exercise XLIX, 
Ex. 12), and his table was a table of chords, not of sines. (See Art. 70.) 
His table gives the values of the chord for each half degree of arc with a 
degree of accuracy somewhat greater than that which would correspond 
to a modern five-place table. The earlier tables of Hipparchus and 
Menelaus are not extant. 

The Hindus followed the method which we have outlined even more 
closely. In fact, the table given by Aryabhata (born 476 a.d.) gives 
the values of the -sine at intervals of 3° 45'. As we have seen, this is 
precisely the interval which would arise as a result of continued bisec- 
tion of 30°. 

Essentially the same method was used in subsequent improvements 
and enlargements of these tables, especially by Rheticus (1514-1574) 
and Pitiscus (1561-1613). Other far more powerful methods have 
since been developed, based essentially on the notions of the calculus 
and the theory of infinite series. 



EXERCISE XLIX 

1. From the functions of 30° find those of 60°. 

2. From the functions of 60° find those of 120°. 

3. From the functions of 90° find those of 45°. 

4. From the functions of 30° find those of 15°. 

5. From the functions of 15° find those of 7°.5. 

6. Find formulae for sin 3 cc, cos 3 a, tan 3 a. 
Hint. Put 3 a = 2 a + a. 

7. Find formulae for sin 4 a, cos 4 a, tan 4 a. 

8. Find formulae for sin 5 a, coS 5 a, tan 5 a. 

9. Prove formula (1), Art. 83, by means of a figure. 

10. Given tan 6 = f, 6 being in the first quadrant. Find the func- 
tions of 2 and \ 6. 

* Ptolemy's great work on Astronomy, usually known as the Almagest, re- 
mained in undisputed authority until the time of Copernicus. The so-called Ptole- 
maic system of astronomy, as opposed to the more modern Copernican system, 
was named after him for this reason. 



CERTAIN LIMIT RELATIONS 193 



•) 



11. If is in the third quadrant and sin $ = — -, find the functions 
of 2 0. V5 

12. In a circle of radius 1, inscribe a regular pentagon. Show that, 
by means of this construction the trigonometric functions of 72 and 18° 
may be computed. In particular, show that 

sin 18° = ](\ 5 - 1). 

13. Making use of the results of Ex. 12, compute the functions of 12 D . 

14. Making use of the results of Ex. 13, compute the functions 
of 8°. 

Assuming the truth of the law of cosines, and setting."? = %(a + b + c), 
prove the following formulae for the functions of the half angles of a 
triangle. 

-ic -ii /(.<? - b) (s -c) 

15. sin \A — \± ^ '- • 

1 be 



he 



17. toM=V C '~* )( '~ C) - 

* s(s - a) 

18. Prove formulae (6) and (7) of Art. 84 by means of equations (4). 

85 a. The limit s ^ 11 — and related limits. Although the method 

sketched in Art. 84 for Circulating a table of the values of 
the trigonometric functions is adequate, it involves far more 
labor than is actually necessary. The following theorem, 
whose truth is almost self-evident, is of great importance in 
this connection as it enables us to calculate the sine of a 
very small angle with a minimum of effort. 

It an angle or arc is expressed in radians, the quotient — — 



approaches the limit 1 when the angle itself approaches zero as 

lit. In symbols 

limit ?*?L? = i. 

*=o 



194 



FUNCTIONS OF MORE THAN ONE ANGLE 




Fig. 104 



But 



In order to prove this theorem, let 
us draw an acute angle A OP = 6 as in 
Fig. 104, and symmetrically the angle 
A OQ also equal to 6. With the vertex 
as center and any convenient radius 
r, draw the circular arc PA Q and join 
PQ, intersecting OA in M. The tan- 
gents PT and QT, at P and ft will 
meet in a point T of OA prolonged. 

Obviously we shall have 
(1) PQ<nvcPAQ<PT+ TQ. 



PQ = 2 PM= 2 r sin 0, 
arc PAQ = 2 arc AP = 2 r<9 (Art. 74, equation (5)), 
P2 r +2 T #=2.P2 T =2rtan0, 

where the truth of the second equation depends essentially 
upon our assumption that 6 is expressed in radians. If these 
values be substituted in (1)^ we find 

2 r sin d < 2 r < 2 r tan 0, 

or, after division by the positive quantity 2 r, 

(2) sin 6 < 6 < tan 6. 

If we divide all three members of this inequality by the 
positive number sin 0, we find 

(3) 1 <—*< —a 

sin o cos c/ 

We know that cos 6 approaches the limit 1 when 6 ap- 

proaches zero. Since the value of — — -, according to (3), 

sin 6 

, which latter quantity itself ap- 



lies between 1 and 



cos 6 



proaches 1, 
(4) 



e 



sin 6 



must also have 1 as its limit. That is, 



limit e 
0=o sin 6 



= 1. 



CERTAIN LIMIT RELATIONS 195 

From (3) we have further 

., . sin . n 

1 > —^- > cos 0, 

u 
so that, by a similar argument, we find 

:>) limit si-e 

^ ' 0=0 

If we divide all members of (2) by the positive quantity 
tan 0, we find 

COS0< -< 1, 

tan 
SO that 

( ,;) limit A = limit tan ^ = i 

v «=° tan «=° 

Since 

sin ( — 0) _ — sin _ sin 
-0 ~~^T ~0~' 
tan(- 0) _ - tan _ tan 
-0 -0 e ' " 

equations (4), (5), and (6) will still remain true if ap- 
proaches zero through negative instead of positive values. 

These forrnuke have many important applications. For 
the present, we shall mention only the one indicated at the 
beginning of this article. The content of equation (5) may 
be formulated as follows. If we w T rite 

r*r\ sin 1 rv 

(O ^ = 1 - s < 

the angle (expressed in radians) may be chosen so small 
that B [the difference between 1 and ■ ) will become less 

than any previously assigned quantity. Since we find 

fr0m(7) Bin*-*— «*, 

we see that w^e can make the angle (expressed in radians) 
so small that the difference between and sin becomes less 
than any previously assigned small fraction of 0. 



196 FUNCTIONS OF MORE THAN ONE ANGLE 

Suppose, for instance, that we wish to compute the sine of 
a small angle to 5 decimal places, that is, with an error which 
shall be less than 5 units of the sixth decimal place, or 
.000005. We now know that the angle may be chosen so 
small that its sine may be equated to the radian measure of 
the angle itself with an error of less than 5 units of the sixth 
decimal place. In other words, the equation 

(8) sin 0=0 (in radians) 

will be true up to five decimal places for all angles which are 
sufficiently small. 

Of course, our method does not inform us just how small 
must be in order that equation (8) may be true up to five 
decimal places. It would take us too far afield to investigate 
this question, a complete answer to which is beyond the scope 
of this book. The student may convince himself, however, 
by actual comparison with the tables, that equation (8) is 
true to five decimal places for all angles less than 2°. In all 
numerical work, then, involving such small angles, no error 
noticeable in five-place calculations is introduced by putting 
sin 0=0 (in radians). 

Since we have 

cos = 1 - 2 sin 2 - (Art. 84, equation (1)), 

we may put 

(9) cos0=l-2(£f = l-^, 

a formula which will certainly be true up to the fifth decimal 
place for all angles less than 2°. In fact equation (9) holds to 
five decimal places even for angles much larger than this and 
may serve for the purpose of computing the cosines of such 
angles. Since 2 is small as compared with 0, if itself is 
small, we shall even be justified in equating cos to 1 for 
very small angles. Our tables show that no error is intro- 
duced in five-place calculations by putting cos = 1, if is 
less than 0° 16'. 



CERTAIN LIMIT RELATIONS 



197 



Since (8) is true bo five decimal places if 6 < 2°, we see 
that we shall have 

sin 26 = 26 

with the same degree of approximation if 6 < 1°. More 
generally the formula 

(10) sin nO = nO (# in radians) 

is correct to live decimal places it" n6 is less than 2°. 

The results deduced in this article make it very easy to 
compute the functions of very small angles. By combining 
these results with the methods of Art. 83 an extensive table 
of the trigonometric functions may be constructed with com- 
parative ease. 

EXERCISE L 

Compute the values of the following functions of small angles to five 
decimal places by the method of Art. 85 a and compare with the values 
obtained from the table: 



sin 12'. 
tan 15'. 



3. sin 1°. 

4. cosl°. 



5. tan 1°. 

6. cotl . 



7. What will be the angle subtended by a lamp-post 10 feet high at a 
distance of one mile? 

8. In order to find the distance from the earth to the moon, the 
following plan may be adopted. Two astronomers stationed at A and l'> 
respectively (Fig. 105) observe 

at the same instant the angular 
distance of the moon's center M 
from their respective zeniths 
(their overhead points), Z and 
Z '. This gives the angles 

B= ZAM and f3= Z'BM. 
For the sake of simplicity as- 
sume that both stations .1 and 

B are on the equator, that the moon is in the plane of the equator, and 
let E be the center of the earth. Then Z A EB = A is equal to th<- differ- 
ence between the longitudes of the two stations and may be regarded as 
known. We may now compute 

Z EA M = 180° - a, Z EBM = 180° -ft Z . 1 /•;/>' = A. 




Fig. 105 



198 FUNCTIONS OF MORE THAN ONE ANGLE 

Now the sum of the four angles of the quadrilateral AEBM is four right 
angles ; that is, 

ZM+ 360° -a - (3+ X = 360°, 
whence 

ZM=a + p-\. 

From the value of M obtained in this way, it is easy to compute the 
angle subtended by the earth's radius at the center of the moon. This 
angle is called the moon's parallax. 

Find the moon's distance from the earth if the moon's parallax is 57' 
and if the earth's radius is 4000 miles. 

9. The apparent diameter of the moon as seen from the earth is 
about 31'. Making use of the result of Ex. 8, what is the moon's 
diameter in miles? 

10. The sun's parallax is about 8". 8. Assuming 4000 miles as the 
length of the earth's radius, find the distance from the earth to the sun. 

85 b. The auxiliary quantities S and T. We have seen in 
Art. 28 that the ordinary tables of sines and tangents be- 
come inconvenient for very small angles. To avoid this in- 
convenience, we constructed an additional table (Table III), 
giving the values of the sines and tangents of such small 
angles directly for every second of arc. But we may accom- 
plish the same purpose in another way, by means of an 
auxiliary table occupying far less space than the additional 
table just mentioned. This second method is based on the 
fact that the quotients 

sin 6 -, tan 

nr and — 

change very slowly if 6 is a small angle. 

We have just seen that each of these quotients has unity 
as its limit when 6 approaches zero, provided that the angle 
is measured in radians. Let us instead express 6 in minutes 
of arc. Let 6' denote the number of minutes and 6^ R) the 
number of radians contained in the angle 6. Then, accord- 
ing to Art. 85 a, 

d) limit ™Li = limit *5Hi=l. 



AUXILIARY QUANTITIES S AND T 199 

Since (Art. 74) 

1° = -- radians = 0.0174588 radian, 
180 

and therefore r = 0.0002909 radian, 

the angle 0, which contains 0' minutes, will contain 

&n = 0.0002909 0' radian. 
Consequently we lind 

^ = sin 0_, _*? = 0.0002909 ?!"' 

0' 0.0002909 0<*> ' 

and therefore, on account of (1), 

(2) limit !1B_^ = limit tan(9 = 0.0002909. 

e±o 0' e±o 0' 

In other words, if 0' is an angle expressed in minutes, the 

common limit toward which — — — and — tend, when 6' 

d 

approaches zero, is a number whose first seven decimal places 

are given by 0.0002909. 

Let us write 

, N sin . tan 6 

(3) •-- jr, *—§r- 

These quantities change their values very slowly for small 
values of 0. In fact we have just seen that, for angles which 
are small enough, we shall have 

log s = log t = log 0.0002909 = 6.46373 - 10. 

o o o 

For e = 2° = 120' we have 

sin 2° . tan 2° 

8 = , t = — •, 

120 120 ' 

which gives, if we look up the logarithms from the tables, 

log sin 2° = 8.64282- 10 log tan 2°= 8.54308 - 10 

Log 120 = 2.07918 log 120 = 2.07918 



log s = 6.46364 - 10 Log t = 0.46890 - 10 

Therefore, while changes from 0° to 2°, log 8 changes only 
by 9 units and log t by 17 units of the fifth decimal place. 



200 FUNCTIONS OF MORE THAN ONE ANGLE 

Table IV enables us to find the values of S = log s and 
T=logt for every angle between 0° and 2°. To find the 
logarithm of the sine or tangent of such an angle we have 
the formulae (immediate consequences of (3)) 

.,. log sin 6 = log 6' + log s = log 0' + S, 

log tan = log & + log t = log 6 f + T. 

If the log sin or log tan of a small angle is given, to find 
the angle, we may do this by means of one of the equations 

.rx log 6' = log sin 6 — S, 

log 6< = log tan 6 - T, 

obtained from (4) by transposition. 

Of course the quantities S and T are available, not only 
for sines and tangents- of small angles, but also for cosines 
and cotangents of angles close to 90°. 

EXERCISE LI 

1. Find the sine and tangent of 1° 13'. 21 by using the auxiliaries S 
and T. 

Solution. Since = 1° 13'. 21, we have $' = 73'. 21. 

log 0' = 1.86457 log 0' = 1.86457 

S = 6.46369 - 10 T = 6.46379 - 10 

log sin = 8.32826 - 10 log tan = 8.32836 - 10 

2. Given log sin = 8.24798 - 10. Find 0. 

Solution. We find from Table IV corresponding to log sin 0=8.24798 
- 10, S = 6.46370. Formula (5) leads to the calculation 
log sin = 8.24798 - 10 
S = 6.46370 - 10 
log 6' = 1.78428 .-.0' = 60'.85 = 1° 0'.85. 

Find the values of the logarithms of the following functions by means 
of the auxiliaries S and T: 

3. sin 1° 21'.63. 5. cos 89° 13'.21. 

4. tan0°32'.61. 6. cot 88° 21'.75. 

Find the angles determined by the following functions by means of 
5 and T: 

7. log sin 6 = 7.76345 - 10. 9. log cos = 8.42371 - 10. 

8. log tan = 8.50731 - 10. 10. log cot = 8.53729 - 10. 



CHAPTER XII 

DIRECTED LINES AND DIRECTED LINE-SEGMENTS * 

86- Plan of another proof for the addition formulae. When 
we proved the addition theorem in Art. 79, we found it 
necessary to divide the proof into a number of cases accord- 
ing as the angles were in the first, second, third, or fourth 
quadrants. To be sure, by making use of the method of 
mathematical induction we found it a fairly simple matter to 
make an exhaustive discussion covering all cases. Neverthe- 
less we feel that it must be possible to devise a method 
enabling us to prove this theorem at one stroke for angles of 
any magnitude. The key to the solution of this problem is 
found to be a careful formulation of the notions of a directed 
line and a directed line- segment. Since these notions are of 
very great importance, not only in this connection, but in 
many other parts of pure and applied mathematics, we shall 
tind it worth our while to speak of them, even if they are 
not absolutely indispensable for the proof of the addition 
theorem. 

87. Directed lines and segments. A straight line is infi- 
nite in extent and is determined by any two distinct points 
upon it. We may, however, think of one and the same 
straight line as having either of two opposite directions, in 
which case we speak of it as a directed line. Since we can 
never draw more than a finite portion of a line, we may 
indicate the direction of a directed line by placing a + sign 
near one end of that portion which actually appears in the 
figure. In Y\^. 106 we have thus indicated tin- direction of 

* This chapter may be omitted in a first course it" the time is insufficient. 

201 



202 



DIRECTED LINES AND LINE-SEGMENTS 




the directed line I which is to be thought of as pointing 
toward the upper right-hand corner of the page. 

This method of indicating the direction of a di- 
rected line has already been used in this book to 
indicate the positive direction of the x-axis and #-axis 
of a system of rectangular coordinates. (See Art. 63.) 
These are directed lines. 

A line-segment is a finite portion of a line 
and may be described by naming its end points, such as AB 
in Fig. 106. But again, we may think of it as a directed 
line-segment, thus distinguishing between AB and BA. 

When a directed line-segment lies upon a directed line, its 
direction or sense may be the same as that of the line or else 
opposite to it. If a directed line-segment on I is 5 units 
long and if its direction is the same as that of I, we may 
represent it by the number -f- 5. A line-segment of the 
same length and opposite direction will be represented by — 5. 
In general, a directed line-segment, which lies on a directed 
line, shall be counted positive or negative according as it has the 
same or the opposite direction as the directed line. For such 
line-segments, we always have BA = —AB, or AB + BA = 0. 

We are now ready to prove the following theorem. If A, 
B, C, are any three points on a directed line, then 

(1) AB + BC=AC, 

where AB, B C, and A C are directed line-segments. 

Proof. 1. Let A C be positive and let B 
be between A and C. Then AB and BO are 
also positive and the truth of the theorem, in 
this case, is obvious. (See Fig. 107.) 

2. Let A C be positive, but let C be be- 
tween A and B. Then AB is positive, but BC 
is negative and equal to — OB. Thus 
AB + BC = AB - CB = AC. (See Fig. 108.) 

3. Let AC still be positive, but let A be be- 
tween B and C. Then (Fig. 109), 

AB + BC= -BA + BC= BC-BA = AC. 




Fig. 107 



By 




Fig. 108 




DIRFXTED LINKS AND SEGMENTS 203 

If AOis negative, there are again three eases according as 
the order of the three points is CBA. BOA, or CAB. But 
the above relation (1) will be found to be true in all ci 
These last three cases may, of course, also be reduced to the 
former three by reversing the direction of the line /. 

The following is a simple corollary of the above theorem. 
It A. B. C D. are any four points of a directed line, tve have 
the relation 

(2) AB+BC+CD = AD 

between the directed line-segments AB. BC, CI), and AD. 

In faet. by the theorem just proved, we have 

AB+BC=AC AC + CD = AD, 

so that we find by addition 

AB + BC+ AC+ CD = AC + AD, 

which reduces to (2) if we subtract AC from both members. 

It may now be proved by induction that, in general, if 
A. B. C ••• 31 N are any finite number of points on a directed 
line, then 

.:')) AB + BC+CD+ .. +MN=AN. 

Equations (1), (2), (3) may also be proved by algebra. 
On the directed line I, let us introduce a point as origin 
or zero point of a scale, whose positive readings are on that 
side of which corresponds to the positive direction of the 
line /. This is precisely what we did when we established 
scales upon the z-axis and y-axis of a 
coordinate system (Art. (33). Denote 
by l A the leading of the scale which 
corresponds t<> the point A. by l B that 
which corresponds to B. The differ- 

ence 1 ' B — I , will give the length of the line-segment AB, 
affected with a plus or minus sign according as AB is a 
positive or negative line-segment in the sense of our pre- 
vious definition. 



204 



DIRECTED LINES AND LINE-SEGMENTS 




If then we have any three points A, J9, C on the directed 
line, we shall have 

A.Jj = L B — l A , £> C = L c — Ifr AC = Iq — l A , 

and therefore 

AB + BC = l B -l A + l c -l B =l c -l A = AC, 
which is the same as (1). In the same way we may also 
prove equations (2) and (3). 

88. Angles between directed lines. Let Z and m be two 

directed lines, and let us denote the angle between their 
positive directions by (Z, m) or (ra, Z) according as we think 
of Z or m as the initial side of the angle. To be perfectly 
specific, we understand by (Z, m) the angle, less than 360°, 
^ through which it would be necessary to 

rotate the directed line Z in the counter- 
clockwise direction, in order to make its 
positive direction coincide with the posi- 
tive direction of the directed line m. In 
Fig. Ill, the angle (Z, m) is marked 0. 
Similarly (w, I) is the angle through 
which m would have to be turned in the positive (counter- 
clockwise) direction in order to make the positive direction 
of m coincide with that of I. In Fig. Ill (w, I) is marked 0' '. 
We see that we shall always have 

(1) (7, m) + (m, = 360°, 
so that 

O, l)= 360°- (Z, m) 

and therefore (see Art. 77, equations (7)), 

(2) sin (m, T)= — sin (Z, w), cos (w, Z)= cos (Z, m). 

89. Projections. The projection of a point P on a line Z 
is the foot of the perpendicular dropped from the point to 
the line. The projection of a line-segment 
AB on a line I (see Fig. 112) is the line-seg- 
ment A'B' of Z bounded by the projections of 
A and B. If AB is a directed line-segment, so ^ 
is A'B'. Fig. 112 



Fig. Ill 



PROJECTIONS 



205 



We wish to solve the following problem. Given a directed 
line-segment AB on a directed line p; to find the magnitude 
and sign of its projection upon any second directed line I. 

The Line-segment AB may have the same sense as the 
directed line p or else the opposite sense: it will be repre- 
sented by a positive number in the first ease and by a nega- 
tive number in the second (Art. 87). If we denote by 
AB the positive number which represents merely the 
length (regardless of direction) of the line-segment AB, 
we shall have 

(1) AB = | AB , read AB = length AB, 
or 

(2) AB = - AB . read AB = minus length AB, 
according as the direction of AB agrees with that of the 
directed line p or not. 

Let us consider first the case (1) in which AB is positive. 
(See Figs. 118 and 114.) Let OM he the projection of AB 
on /. Choose as origin and I as the z-axis of a system of 




I'm. 11 



*P\ 


B 


+ 1 


I 


+i '\ 


J- 




A 




M 













X! 



Fig. 114 



coordinates, so that the positive z-axis coincides with the 
positive direction of the line I. The line m through (9, per- 
pendicular to / and with its positive direction as indicated in 
the figures, will then be the y-axis. Through draw a 
directed line p' parallel to p and let P be its intersection 
with the line MB through B perpendicular to L* The line- 

*The word parallel in this theory means, not only that the lines arc parallel 
in the ordinary Bense, hut that their positive directions arc the same. The word 
anti-parallel is sometimes used for two parallel directed lines whose positive 
directions are oppi 



206 



DIRECTED LINES AND LINE-SEGMENTS 



segments AB and OP have the same length and direction, 
and the same directed line-segment OM on I as projection. 
But, by the definition of the cosine of a general angle 
(Art. 64), „ ^ „ ^ 0M 



cos (I, p) = cos (I, p'} 



OP' 



since OM is the abscissa and OP the radius vector of a point 
P on the terminal side of the angle (7, p'~), this angle being 
in its standard position and OP = AB being positive in the 
case under consideration. Consequently we have 

OM = OP cos (I, p) = AB cos (I, p). 

Since OM is the projection of AB on I, we may write this 
as follows : 

(3) OM = projj AB = AB cos (Z, jt?) = AB cos (jt>, ;* 

for, according to equation (2), Art. 88, the angles (7, p~) and 
(p, £) have the same cosine. 

Thus the projection upon the directed line I, of the positive 
line-segment AB of the directed line p; is equal to AB mul- 
tiplied by the cosine of the angle between I and p. The 
projection is a directed line-segment on I, positive if (7, p) is 
+ p in the first or fourth, negative if 
(7, />) is in the second or third 
quadrant. 

We have proved equation (3) 
when AB is positive. Let us con- 
sider now the case when AB is a' 
negative line-segment of p. (See 
Fig. 115.) The projection of AB, 
as well as that of OP, is OM. 
We may think of OP, which is a negative line-segment on 
p', as a positive line-segment of a directed line p", cokicident 
with p' but having the opposite direction. But this latter 
directed line makes an angle with I just 180° greater than 
(I, p f ) ; that is, 

(4) (l,p") = (*,/) + 180°. 



+ m 




Fig. 115 



* The symbol proj^ AB is read " projection of AB upon 



I" 



PROJECTION OF A BROKEN LINK 207 

For if we rotate I in the counterclockwise direction around 
as center, it will take 180° more to make -f / coincide with 
-f p" than with + p' . According to formula (8), which we 
know to he valid for all positive line-segments, we have 

therefore , _ . _ 

proj, AB = \AB\ cos (/./'), 

where AB \ denotes the length of the line-segment AB taken 
as a positive number. But according to (4) and equations 
(4) o( Art. 77. we have 

cos (7, _/>") = — cos (7, p') == — cos (7, p~) = — cos (^, Z), 
so that 

proj, AB = - | AB | cos (7, p~) =-\AB\ cos (>, 0- 

Since in our case AB was a negative line-segment, we had 
(cf. equation (2)) 

^LS = - | AB |, 

so that we may write finally 

(3) proj j AB — AB cos (7, p) = AB cos (p, I). 

In other words, formula (2>} for the projection upon a directed 
line L of a directed line-segment AB of a second directed line 
j>. is true for both positive and negative line- segments. 

If we think of the line-segments as mere lengths, not 
endowed with direction, formula (3) may be simplified to 

A'£' = proj|AB = ABcos0, b^ ^b 

where 6 is the angle between AB 



and the line Z, this angle being tin- B A 

, . ,1 . . I % i , Fig. 116 Fig. 117 

derstood m the sense ot elementary 

geometry without any reference to a positive sense of rota- 
tion. Consequently 6 may be regarded as an acute angle, 
so that its cosine will always be positive. (See Figs, lib 
and 117.) 

90. Projection of a broken line. Let us connect two points 
A and Oby a directed line-segment AC. We may think of 



208 



DIRECTED LINES AND LINE-SEGMENTS 



this line-segment as describing the shortest path from A to 
O in magnitude and direction. Let ABO be a second 
b (longer) path from A to O made up of 

o two directed line-segments AB and BO 
(Fig. 118). 

Let us project these three line-segments 




A! 



B' C 
Fig. 118 



H 



on any directed line I, so that 

(1) A'B' = proj, AB, B< 0< = proj, BO, A' C = proj, AC. 

Since A'B', B' 0' and A' 0' are three directed line-segments 
of a directed line, we have (Art. 87, equation (1)) 

A' 0' = A'B' + B'C; 

whence, substituting the values (1), 

(2) projj AC= proj, AB + proj, BO. 

By means of the more general relation (3) of Art. 87, we 
may prove the following general theorem, of which (2) ex- 
presses the simplest special case. 

Let us connect two points by two paths, one of which is a 
single directed line-segment, while the other is made up of a 
finite number of directed line -segments. Then the projection, 
upon any directed line, of the first path is equal to the sum of 
the projections of the various 
line-segments which make up 
the second path. 

Figure 119, in which the va- 
rious line -segments are marked 
with arrowheads to indicate their 
directions, illustrates this theo- 
rem. Since the positive direc- 
tion of I is toward the right, the projections A'G', A'B', B'C, D'E', 
E'F>, F'G' are all positive, while the projection CD' of CD is negative. 
Consequently 

projj AB + projj BC + projj CD 

= A'B' + B'C + CD' = A'B' + B'C - D'C = A'D'- 

Let the line I in Fig. 118 coincide with AC, and denote by 
a, b, c the lengths of the three line-segments BO, AO, and 




B' D' C E' 
Fig. 119 



DIRECTION COSINES OF A LINE 



209 




AB respectively. (See Fig. 120.) 
A cording to (2) and Art. 89, equa- 
tion (J3). we shall bave 

(3) b = c cos A + a qo* (7, 
a relation which may be used to ad- 
vantage in many problems concerning triangles. Of course 
the two equations similar to (3) 
, . c = a cos B -\- b cos A, 

a = b cos C + c cos B, 
may be proved in the same fashion. 



91. Direction cosines of a line. Let V be a directed line 
through the origin of coordinates. The angles (z, I') and 
(/', ^/), which its positive direction makes with the positive 
.r-axis and ^-axis, are called its direction angles. The direc- 
tion angles (x, I) and (7, #) of a line I which does not pass 
through the origin are defined to be the same as those of a 
parallel line V which does pass through the origin. Observe 
that the angle (a:, I) has the z-axis as initial side, while the 
initial side of the second direction angle (7, ?/) is not the 
^-axis, but the line I. 

If the angle (z, I) is in the first quadrant (Fig. 121), we 
clearly have 
(1) (x, + (^)=90°. 





Fig. 121 



Fig. 122 



If the angle (x. I) is in the second quadrant (Fig. 122), 
wehave ', = (,-. 0=00° + «. 



a. 



210 



DIRECTED LINES AND LINE-SEGMENTS 



For clearly it requires a positive (counterclockwise) rotation 
of 360° — a to make + V coincide with -f y. Therefore we 
have, in this case, 

(2) 0M) + O,y) = 360° + 90 o , 

and the same relation holds if (#, Z) is in the third or fourth 
quadrant, as may be verified easily. Thus we always have 
either (Z, y)=90°-"(as, V) or (Z, y) = 360° + 90° - <>, Z), so 
that in all cases 

(3) cos (j/, Z) = cos (Z, y) = sin (x, Z), sin (Z, y) = cos (#, I). 
The two quantities cos (a;, Z) and cos(?/, Z) are called the 
direction cosines of the directed line I. 

Since we have, for any angle, 

cos 2 (#, Z) + sin 2 (#, Z) = 1, 
we find from (3) the following simple relation between the 
direction cosines of a line I ; 
(2) cos 2 (x, I) + cos 2 (y, I) = 1. 

92, Formula for the cosine of the angle between two lines 
whose direction cosines are given. Let us consider two 
directed lines Z and m whose direction cosines are cos (#, Z), 
cos(y, Z), and cos (#, m), cos(y, m) respectively. We wish 
to find a formula for the cosine of the angle between the 
two lines. 

We may assume that the lines Z and m pass through the 
origin. If they do not, we may first solve the problem for 

two lines Z', m\ parallel to Z, m re- 
spectively, which do pass through 
the origin. Since-?', m' have the 
same direction cosines as Z and w?, 
and since the angle between Z', m f is 
equal to that between Z, m, the two 
problems are clearly equivalent. 

Let us choose a point M (see 
Fig. 123), on the line w, such that 
the line-segment OM is positive. Let X be the projection 
of M on the #-axis. Then the projection of the broken 




Fig. 123 



COSINE OF ANGLE BETWEEN TWO LINKS :>11 

line 0X31 on 1 will be equal to that of OM (Art. DO, equa- 
tion (^-)). That is. 

(1) projj 031= projj 0X + projj XM. 

On account of Art. 89, equation (3), we have 

^ proj z (W=ai/cos(>», I), 
^ L) proj, 0X= OX vos {x, /), proj, XJ/= XMoos(y, /> 
since XJ/ is a directed line-segment on a directed line y' 
parallel to the //-axis.* 

Substitution of (2) in (1) gives 

(3) OMcosQn, /) = 0Xcos(x, l)+XMcos(y, I). 

But <9A" = proj, 03f = OM cos (a, m), 

JOf =projV 0ilf= proj„01T= OMcos(y, rn), 

since the projections of 6W, on the two parallel directed 
lines y and y'. are equal. If we substitute these values in 
(3), we find 
0ifcos(m, 

= 031 cos (z, I) cos (#, m) -f OM cos (?/, Z) cos (y, w) 
or, upon division by OM. 

(4) cos (m, J) = cos (x, 2) cos (a?, in) + cos (?/, £) cos (i/, m), 

the desired formula. 

The proof which we have given of this formula is perfectly 
general ; that is, it is applicable, no matter how large or 
small the angles (7, m), (#, Z), etc., may be, in what quadrants 
they happen to lie, or whether they are positive or negative. 
In fact, the figure (Fig. 123) has not really been used in the 
demonstration except for the purpose of suggesting the order 
of the various steps of the argument. Every step of this 
proof can be justified by quoting a previously demonstrated 
general theorem. 

It will he a good exercise for the student to repeat the argument with 
a different figure in which some or all of the angles concerned are not 
acate. 

* In Fig. 123 we have chosen the positive direction ot y' to correspond to thai 

of y. This is Dot essential, but it is convenient. 



212 DIRECTED LINES AND LINE-SEGMENTS 

93. New proof for the addition and subtraction formulae. 

Let us denote the angles (x, V) and (x, m) by a and 
respectively. Then 

cos (7, m) — cos (a — /3), 
and (Art. 91, equations (3)), 

cos (j/, 1) = sin (re, Z) = sin a, cos (j/, m) = sin (#, m) = sin fi. 
Consequently we find, from equation (4) of Art. 92, 
(1) cos (a — yS) = cos a cos /3 -f sin a sin /3, 

the subtraction formula for the cosine. We shall leave it as 
an exercise for the student to deduce from this the subtrac- 
tion formula for the sine and the addition formulse for both 
functions. (See Exercise LII.) 

94. The generalized law of sines. If we attribute a 
definite direction to every line of the plane, and define angles 
between them, as in Art. 88, with reference to a positive 
direction of rotation, the angle between any two such directed 
lines may be greater than 180° and the trigonometric func- 
tions of such angles will have definite signs as well as numer- 
ical magnitude. Moreover, every line-segment will then also 
have a definite sign. 

It may be shown that the law of sines, when written in the 
form 

sin (b, c) sin (<?, a) sin (a, b) 

will be true of any triangle, not merely with regard to the mag- 
nitudes of the quantities involved, but also with regard to their 
signs. 

In order to explain this statement, con- 
sider first the case that the positive direc- 
tions of the three directed lines a, b, c 
(the sides of the triangle) are from B 
toward C, from C toward A, and from 
A toward B respectively. (Cf . Fig. 124.) 
Fig. 124 Then BC, CA, and AB are all positive, 




GENERALIZED LAW OF SINES 213 

and the three denominators in (1) are either all positive or 
all negative. 

In Fig, 124 they sure all positive. But bhey would all be negative if 
the clockwise rotation had been chosen as positive direct ion of rotation. 
They would all ho negative, even without any change in tin choice of the 
positive direction of rotation, if the points named .1 and />' in Fig - . 124 
were interchanged. 

Consequently equations (1) are true in this case, not 
merely numerically, but also with regard to sign. 

Let us now invert the positive direction of a single one of 
the lines, that of a, for instance. Then BChecomes negative; 
the angle (£, c) and the segments CA and AB remain unal- 
tered ; (c, a ) and (a, b) each change by 180°, so that their sines 
change sign. Consequently equations (1) will, still be verified. 

By combining several such changes we easily arrive at the 
conclusion that equations (1) will always be true, in magni- 
tude and sign, no matter how the positive directions of the 
three lines a. 6, e may have been selected or which of the two 
opposite kinds of rotation be regarded as positive. 

This generalization of the law of sines is due to the great geometer 
Mobius (1790-1868), and is of great importance in many applications, 
especially in projective geometry. 

EXERCISE Lll 

1. Show that the principal theorem of Art. 90 may be enunciated as 
follows. If a finite number of directed line-segments form a closed poly- 
gon, the sum of their projections upon any directed line is equal to zero. 

2. Show that the ./-component of the resultant of two forces (cf. 
Art. 58) is equal to the sum of the ^-components of the two original 
force.-. Similarly for the //-components. 

3. From formula (1) of Art. 9:> deduce the formula for sin (a — (3). 

4. From formula (1) of Art. 93 deduce the formulae for cos (« + (3) 
and Bin (a + (3). 

5. Generalize the law of tangents from the standpoint of directed 
Line-segments in a way which shall be analogous to the generalization of 
the law of sines carried oul in Art. 94. 

Hint. The law of tangents may be deduced from the law of sines. 

6. Generalize in similar fashion the projection formulas ('■'<) and (4) 
Of Art. 90. 



CHAPTER XIII 

THE INVERSE TRIGONOMETRIC FUNCTIONS AND 
TRIGONOMETRIC EQUATIONS 

95. The problem of inverting the trigonometric functions. 

In the light of our recent studies we may say, with a con- 
siderable degree of propriety, that trigonometry is a discus- 
sion of the properties of the trigonometric functions. However, 
our discussion of these functions, so far, has been somewhat 
one-sided. With the exception of a few practical applica- 
tions in the first part of the book, we have always looked 
upon these functions from the point of view of what may 
be called the direct problem ; that is, given the angle, to find 
the function. We now propose to discuss, somewhat more 
fully than has been done so far, the inverse problem ; that is, 
given the value of one of the functions, to find the corre- 
sponding angles. 

We notice at once a fundamental difference between these 
two problems, which may be illustrated by the relation be- 
tween an angle and its sine. Every angle has only one sine, 
but there are many angles which have the same sine. Con- 
sequently while the direct problem (to find the sine of a given 
angle) has only one solution, the inverse problem (to find an 
angle with a given sine) has many solutions. 

96. Determination of all of the angles which correspond to 
a given value of one of the functions. When we are solving 
triangles, the angles are necessarily either in the first or second 
quadrant, so that in most cases we experience little difficulty 
in finding a unique angle as an answer to a given problem 
of this kind. But, even in this restricted field, we found 
that a problem may have two solutions, owing to the fact that 

214 



FINDING ALL ANGLES FOR A GIVEN FUNCTION 21f> 

there exists two angles 0. one acute and one obtuse, corre- 
sponding to a given positive value of sin 0. (Cf. Art. 55.} 

If the sine of an angle is given, and no particular quad- 
rant or sot of quadrants is prescribed for the angle, the 
number of values which the angle may have is unlimited. 

We may {trove this and find out how all of these angles 
arc related to each other as follows: 

Let the given value of the sine be denoted by s. If s is 
numerically greater than unity, the problem has no solution, 
since no angle has a sine numerically greater than 1. If s is 
numerically less than 1 and positive, there exists an acute 
angle and an obtuse angle 180° — 0, such that 

sin = sin (180°- 0)= s. 

But, on account of the periodic character of the sine, we also 

have sin (0 + n • 860°) = sin = s, 

sin (180° -(9+ n ■ 360°)= sin (180° - 0) = s, 

where n is any positive or negative integer or zero. We see 
therefore that all angles, such as 

(1) n- 360° + = 2 n- 180° + 0, 

or 

(2) n • 360° + 180° - = (2 n + 1) • 180° - 0, 

have the same sine as the angle 0. This maybe expressed as 
follows. All of those angles which can be obtained by add- 
ing a given angle to any even multiple of 180° or else by 
subtracting the given angle from any odd multiple of 180°, 
have the same sine. 

That these are the only angles which have the same sine 
follows easily from the fact that two distinct angles in the 
same quadrant cannot have the same sine. 

If the given value of s is negative, nothing essential is 
changed in the above argument, except that will then be in 
the third or fourth quadrant instead of being an acute 
angle. It will still be true that all of the angles given by 
formula;- (1) and (2j, and no others, have the same sine as 0. 



216 INVERSE TRIGONOMETRIC FUNCTIONS 

Now we may include all of the angles (1) and (2) in the 
single expression 

(3) m ■ 180° ± 0, 

where m may be any integer, even or odd, and where the + 
or — sign is to be used according as m is even or odd. We 
may avoid this awkward distinction by writing in place of (3) 

(4) w .18O° + (-l)™0, 

since (— l) m is equal to + 1 or — 1 according as m is even 
or odd. 

Thus, if is one angle whose sine is equal to a given number 
s, the most general angle which has the same sine is 

m • 180° + ( - V) m 0, 

where m is any positive or negative integer or zero. 

A brief way of indicating this fact is given by the follow- 
ing equations: 

(5) sin \m • 180° + ( - 1 )*» 0] = sin 
or 

sin [mir + ( - l) m 0] = sin 0, 

where we use the first or the second form according as 6 is 
expressed in degrees or in radians. Since the cosecant of an 
angle is the reciprocal of its sine, we have also 

(6) esc [m • 180° -f- ( - l) m 0] = esc 6 
or 

esc [mir + ( — 1)™ 0] = esc 6. 

We find, by precisely similar considerations, 

cos (2 m • 180° ± 0) = cos 0, cos (2 mi? ± (9) = cos 0, 

( 7 ) sec (2 m • 180° ± 0) = sec (9, 0r sec (2 mir ±6)= sec (9, 
and 

tan O • 180° + 6) = tan 0, or tan (mir + 6) = tan 0, 

( 8 ) cot (m • 180° + 0) = cot 0, cot (mir + 0) = cot 0, 

according as the angles are expressed in degrees or in radians. 



DTVERSE TRIGONOMETRIC FUNCTIONS 217 

EXERCISE Llll 

Without making use of tin' trigonometric tables, find all of the angles 
which satisfy the following conditions: 

1. sin 0=]. 3. tan 0=1. 5. cot0 = V3. 7. tan0 = oo. 

2. cos = - \. 4. Bee = V 2. 6. csc 0=1. 8. sin 6 = 0. 

Making use of the tables, find all of the angles which satisfy the fol- 
lowing equations : 

9. Bin 0=*- .4721. 11. tan 0=1.7269. 

10. eos0 = + .3216. 12. sec = 2.7213. 

97. The inverse trigonometric functions. In the equation 

(1) x = sin y, 

we now propose to regard */, the angle or arc, as a function 
of x, the sine. We may express this new way of looking 
at the relation (1 ), by saying that 

y is an angle whose sine is equal to x 
or 

(2) y is an arc tchose sine is equal to x, 

a statement which is usually written in. the contracted form 

(3) y = arc sin x. 

It should be noted that (1) and (3) are merely different 
ways of expressing the same relation between x and y. 
They differ only in one respect. In (1) y is regarded as 
given and x is to be found, while in (3) x is regarded as 
given and y is to be found. The relation between the func- 
tions (1) and (3), that of being inverses of each other, is 
of the same kind as in the more familiar case of the function 
x = y~. which has as its inverse y = ± V#. 

In the sanu; way we define the equation 

y = arc cos x 

to mean that y is an arc whose cosine is equal to x. There- 
fore this equation is equivalent to 

x =cos y. 



218 INVERSE TRIGONOMETRIC FUNCTIONS 

Similarly, if x = tan y, we write 

y = arc tan a?, 

and in the same way we define the symbols 

arc cot x, arc sec x, arc esc as. 

Let us return to equations (1) and (3). We know that 
the sine of an angle is never numerically greater than 1. 
Consequently we can find no angle whose sine is equal to 
x if x is numerically greater than 1. We may express this 
as follows: 

The function arc sin x is defined only for those values of x 
which are not numerically greater than 1. 

If x is numerically less than unity, there exists not merely 
one angle whose sine is equal to #, but the number of such 
angles is unlimited. (See Art. 96.) If x is positive, one of 
the corresponding angles is a positive acute angle. If x is 
negative, the smallest corresponding positive angle is in the 
third quadrant. But in this case there is a negative acute 
angle whose sine is equal to the given negative value of x. 

Let us use the symbol 

Arc sin x, 

distinguished from arc sin x by the use of the capital letter A, 

to indicate the numerically smallest angle or arc whose sine is 

equal to x. 

The function Arc sin x, like arc sin x, is defined only for 

the values of x between — 1 and + 1 ; that is, for those values 

of x for which . 

-1< *<>1. 

But for every such value of #, Arc sin x has only one value, 
while arc sin x has infinitely many values. For positive 
values of #, Arc sin x is a positive acute angle, and for nega- 
tive values of # it is a negative acute angle. No value of Arc 

sin x ever exceeds the limits ± 90° or ± - radians, so that 
we shall always have 

(5) — — ^ Arc sin x 5^ — . 



[NVERSE TRIGONOMETRIC FUNCTIONS 



219 



We shall henceforth speak of Arc sin x as the principal value 
of arc sinx, and we know from equation (5) of Art. 96 thai 

(6) arc sin x = mir ■+-(-—] )'" Arc sin .r, 

where m is any positive or negative integer or zero. 

In many books the symbol arc sin x is used in the restricted 
sense which we have given to Arc sin x. For some purposes 
the distinction between the two functions are sin x and Are 

sin x is not important. But for certain other questions, a 
careful discussion oi the principal value is the only way to 
avoid hopeless confusion. 

This whole matter will become very clear if we make a. 
graph of the function 

I 3 i y = arc sin x. 

Since this equation between x and y has the same signifi- 
cance as 

(1) X = SU\y, 

we may plot the latter relation instead of (3). But we have 
already studied the graph of the similar equation 

(7) y = sin x (see Arts. 73 and 78), 

and clearly the graph of (1) may be obtained from that of 
(7) by interchanging x and y. In other words the graph of 
(1), or what amounts to the same thing, the \y 

graph of (3), is a sine curve placed in a ver- 
tical position. (See Fig. 125.) 

The graph shows clearly that the function 

y = arc sin x 

is not defined for values of x which are nu- 
merically greater than unity. It also shows 
that for every admissible value of x there are 
an infinite number of values of y, viz.. the 
ordinates of all of the points P v P T P 3 , etc., 
in which a line parallel to the ^-axis, at a distance x, inter- 
sects the curve. 



S 






I'm;. L2£ 



220 INVERSE TRIGONOMETRIC FUNCTIONS 

But the curve which corresponds to the principal value of 
arc sin x, 

y — Arc sin x, 

consists merely of that portion of the graph of arc sin x which 
lies between the points A and B. This part of the curve is 
indicated in Fig. 125 by a heavier line.* 

In a similar way we define a principal value Arc cos x for 
the function arc cos x. The function arc cos x is defined for 
all values of x which are not numerically greater than 1, and 
has infinitely many values for every admissible value of x. It is 
convenient to define as its principal value Arc cos x, the smallest 
positive angle whose cosine is equal to x, so that Arc cos x is 
subject to the following inequalities 

^ Arc cos x < 7r. 

The detailed discussion of these statements is left to the 
student as an exercise. 

It should be remarked that the notations sin 1 as, cos -1 as, 
etc., are also in use for arc sin #, arc cos as, etc. This second 
notation, which is frequently used in other branches of 
mathematics, has the advantage of emphasizing the fact 
that sin x and sin -1 x are inverse functions of each other. 
But it has the disadvantage of colliding with the customary 
notation for exponents, and therefore tends to create con- 
fusion. Thus we usually write sin 2 x for (sin re) 2 , and a' 1 
for 1/a. Thus the symbol sin -1 x might consistently be in- 
terpreted to mean 

1 
— = esc a?, 

sin x 

which is something entirely different from arc sin x. 



* If we had chosen as principal value of arc sin x the smallest positive angle 
whose sine is equal to x, the principal value would not be represented by a con- 
tinuous (unbroken) curve. One part of this curve would be OB, and the other 
part (corresponding to negative values of a;) would be between y = rand ?/ = 3 n-/2. 



IXYKRSE TRIGONOMETRIC FUNCTIONS 221 

EXERCISE LIV 

1. Show that the function y = arc tan x is defined for all values of 
x. Draw the graph of t he function and show thai its principal value 
may be selected subject to the conditions 

^ Arc tan x < tt. 

2. Investigate the function arc cot x in the same way. 

3. Are the functions arc seca: and arc esc x defined for all values of 
.?■? Draw their graphs and choose principal values for these functions. 

Find the values of the following expressions: 

4. Arc sin \. 6. Arc tan 1, arc tan 1. 

5. Arc cos \. 7. Arc cos ($V3), arc cos (\ >/:{). 

By using the table of natural functions compute the values of the 
following quantities : 

8. Arc tan 1.3722 + Arc cos 0.4321. 

9. Arc sin 0.3 125 + Arc cot 1.7264. 

10. Obtain the value of sin (Arc sin f + Arc sin |). 

11. Obtain the value of sin (Arc sin § 4- Arc cos^). 

12. If x and y are positive numbers, less than unity, show that 



sin (Arc sin x + Arc sin y) = x\/l — y 2 + //Vl — x 2 . 
Solution. For abbreviation put 

Arc sin x = u, Arc sin y = u. 
Since x and y are positive numbers, less than unity, u and v will be 
positive acute angles such that 
(1) sin u = x, sin v = y, 

and therefore 



(J) cos u — Vl — x 2 , cos v = vl — y 2 . 

We now find 

sin (Arc sin x + Arc sin //) = sin (u + t>)= sin u cos v -f cos u sin r, 
and therefore, on account of (1) and (2), 

sin (Arc sin x + Arc sin //) = xVl — y 2 + yVl — x 2 . 



13. Show that COS (Arc sin x - Arc sin y) = V\ — x 2 Vl — y 2 + xy 
if O^xgl, Ogyg 1. 

14. Show that Arc -in x + Arc cos x=-if-l<r< + l. 



222 INVERSE TRIGONOMETRIC FUNCTIONS 

15. Prove the formula Arc tan x — Arc tan y = Arc tan 



l + xy 

16. What modifications do the equations of Exs. 12 and 13 undergo 
if negative values of x and y are admitted. Discuss in order the cases 
x < 0, y > ; x > 0, y < ; x < 0, y < 0. 

98. Trigonometric equations. It often happens that angles 
are to be determined by means of equations which they must 
satisfy. Such equations usually contain the trigonometric 
functions of the unknown angle and are then known as 
trigonometric equations. 

Let us confine our attention to the case where one unknown 
angle is to be found as a solution of one equation. Such an 
equation may have one of the following forms. 

I. It contains 6 algebraically, but does not contain the 
trigonometric functions of 6. 

Example. Oi-T = o. 

II. It contains the trigonometric functions of the angle 6 
in algebraic combinations, but does not contain the angle 6 
itself explicitly. 

Examples, sin 2 6 — cos = 0, 2 tan + cot 6 = — 3. 

III. It contains the angle 6 and its trigonometric func- 
tions simultaneously. 

Example. 0-1 sin (9 = -. 

2 2 

Clearly the equations of the first type are merely algebraic 
equations, and their discussion properly belongs to a trea- 
tise on algebra. 

It may be shown that the equations of the second type can 
also be reduced to algebraic equations, although it is often 
easier to effect their solution without so reducing them. 

The equations of the third type will not be considered in 
this book. The solution of such equations is a difficult 
matter and can be accomplished in a satisfactory manner 
only in a few cases. It should be mentioned, however, that 



TRIGONOMETRIC EQUATIONS 223 

the method of graphs usually provides an approximate solu- 
tion for such equations. 

We shall now discuss a few simple examples of equations 
of the second type, in such a way as to illustrate the fact 
that their solution may he reduced to that of algebraic 
equations. 

EXERCISE LV 

1. Solve the equation 2 COS 2 2 —5 + 7 sin x = 0. 

Sedition. We have cos-.r = 1 — .sin- a*. Therefore the given equation 
becomes _ 3 _ o sm a x + 7 sin x _ . 

Consequently, if we put siii£ = s, we obtain the quadratic equation 
for*, 2s 2 - 7«+ 3 = 0. 

The solution of this equation gives 

.s- = 3 or s = I . 
The first solution must be discarded, since .s = sin x cannot be numerically 
greater than unity. The second solution tells us that 

sin x = I, 
BO that .r = 30° and x = 150° are the only positive angles smaller than 
which satisfy the equation. 
In examples 2 to 5. find all positive angles less than 360° which satisfy 
the given equations. This may be done without using any tables. 

2. cos20 + cos0 = - 1. 4. tan 2 = - 2 sin 0. 

3. cot 2 0+ tan = - | V3. 5. sec 2 + esc' 2 = 4. 
In solving the following equations, make use of the tables: 
6. sin x tan x = — 2 V 7. cos x cot x = — §. 
8. Solve the equation cos 3 x = sin 2 x. 

99. The equation a sin 9 + b cos 8 = c. The equations of 
the form 
(1) a sin + cos = c 

may be solved by the method of the preceding article. But, 
unless the numbers a, o, c are especially simple, it is far more 
convenient to proceed as follows : 

We introduce two auxiliary quantities I and Z, such that 

_ I sin Z = '/. 

(V) ZcosZ = 6, 



-f 


/ 




P 






z^"^ 


a 







b 1 


If 



224 INVERSE TRIGONOMETRIC FUNCTIONS 

where, moreover, I is assumed to be positive. That it is 
always possible to find a positive number I and an angle L 

which satisfy equations (2) be- 
comes obvious if we think of a 
and b as the rectangular coordi- 
nates of a point P. (See Fig. 
126.) Equations (2) show that I 
is the radius vector OP of P and 
that L is the angle which OP 
makes with the positive direction 
of the a>axis. Our figure also 

Fig. 126 & 

shows us that 

(3) tanZ = £, Z = +V« 2 + 6 2 . 

b 

Of course these same equations also follow directly from (2) 
without the use of geometry. 

If now we substitute the values (2) for a and b in (1), we 

nnd I (sin L sin 6 +• cos L cos 6) = c, 

or by Art. 79, equation (8), 

(4) I cos (0 - X) = <?. 

Therefore, in order to solve (1) we may first determine I 
and L from (2) and then find 6 from (4). 

EXERCISE LVI 

1. Solve the equation 2.1346 sin - 3.0526 cos = 0.9875. 

Solution. 

a = + 2.1346 (1) 

b = - 3.0526 (2) 

c = + 0.9875 (3) 

log a = log (J sin L) = 0.32932 (4) 

log b = \o g(lcosL)= 0.48467 n (5) 

log tan L = 9.84465 n (7) = (4) -(5) 

L = 145° 2'. 15 (8) 

log sin L = 9.75820 (9) 

log cos L = 9.91355 n (10) 

log / c= 0.57112 (11) = (4) - (9) = (5) - (10) 

logc = 9.99454 (6) 

log cos (6-L)= 9.42342 (12) = (6) - (11) 



TRIGONOMETRIC EQUATIONS 225 

$ X -L= 74 ' 87 '.til (13) 

2 - L = 285°22'.39 (11) 

L = 145° 2M5 (8) 

1= =219 39'.76 (15) = (13) + (8) 

6 2 = 70° 24\54 (16) = (14) + (8) 

Remarks. The numbers in parentheses indicate the order in which 
the results are written down and how some of them are obtained. The 
— 10 has not been written down in the case of negative characteristic-. 
The n which follows the logarithms of //. tan /. and cos L indicates that 
the corresponding numbers arc negative. (See Art. 25.) L is chosen in 
the second quadrant because, / being positive, sin L is positive and cos L 
18 negative. The values B l — L and 0., — L. both obtained from (12), 
are the two values, less than 360°. which — L may have so as to corre- 
spond to the given value (12) of logcos(0 — L), 

Check. By substitution in the original equation, 



log a = 0.82932 
log sin 0, = 9.80500 n 
log (a sin Bi) = 0.13432 n 


log ft = 0.48467 n 
log cos 0, = 9.88639 n 
log (ft cos 0i)= 0.37106 


a sin 6 X 
b cos l 


= - 1.3624 
= + 2.3500 



ft COS 0! = +0. 

c = + 0.9875 

This calculation checks 0\ and therefore also L. The relation between 
'i. : . and L is so simple as to make unnecessary a separate check for 2 . 

2. Solve the equation 

- 8.2471 sin + 5.7469 cos = - 6.8271. 

3. Solve the equation 

2.1725 sin $ - 3.2749 cos = 5.7216. 



CHAPTER XIV 



APPLICATIONS TO THE THEORY OF WAVE MOTION 



100. Simple harmonic motion. When we began to study 
trigonometry, it was for a very practical purpose. We 
wished to find an arithmetical method for solving triangles. 
We accomplished this purpose by means of the trigonometric 
functions and by using tables of the numerical values of these 
functions. Later we generalized the notion of the trigono- 
metric functions more than was strictly necessary for the 
simple problem of solving triangles, and we found it to be an 
interesting task to investigate these trigonometric functions 
and their various properties for their own sake. We shall 
now find that these properties, aside from their theoretical 
interest, have in their turn most important applications. 

The fundamental reason for the great importance of the 
trigonometric functions lies in their periodicity. (Cf! Art. 
67.) Many natural phenomena are periodic in character, 
and whenever the attempt is made to represent such a phe- 
nomenon by a mathematical expression, 
the trigonometric functions are found 
to be indispensable. 

The simplest periodically recurring 
motions are connected with uniformly 
rotating bodies. Let the point P 
(Fig. 127) describe a circular path of 
radius a around the point as center, 
and let us assume that it is moving 
Avith a constant angular velocity of eo 
radians per second. Let us assume 
further that P starts its motion at the time t = from the 
point O, which is so located that Z A 00 is equal to a radians. 

226 




Fig. 127 



SIMPLE HARMONIC MOTION 227 

If P is the position of the point at the time £, that is, t 

seconds after the motion has begun, we shall have 

(1) 6 = cot + a, 

where denotes the angle A OP expressed in radians. The 

point will describe its circular path in counterclockwise or 
clockwise fashion according as a> is positive or negative. 

While the point P is moving in its circular path, its pro- 
jection 31 upOD the .r-axis will oscillate to and fro between 
the points A and A', reaching its greatest speed when at 0, 
gradually slowing down until it reaches A', when it reverses 
the direction of its motion, returns with gradually increasing 
speed to 0, after reaching which point it slows down again 
until it reaches A and again reverses its motion. 

To find an analytic expression for the motion of the point 
3L we observe that 

031= x, 0P=a, A0P = 0, OM = OP cos A OP, 

whence, making use of (1), 

(2) x = a cos (cot + a). 

In the same way we see that iV, the projection of P upon 
the y-axis, moves in accordance with the equation 

(3) y =a sin (cot + a). 

Equations (2) and (•>) are so closely related that it will 
suffice to study one of them. In fact, if we put in (3) 

a = - + a. it becomes (cf. Art. 77, equations (2)), 



y = a sin f cot + a' -\ — ) = a cos (cot + «'), 

which is of the same form as (2). We shall therefore con- 
tin*' our attention to equation (3). 

Since any diameter of the circle may be chosen as y-axis, 
we may express our result as follows. If a point describes a 
circular path "'it/, uniform velocity, its projection upon any 
1 diameter of the circle moves in accordance with an equa- 
tion of the form ('■'>). Such a motion is called a simple har- 
monic motion. The quantity a which measures the maximum 



228 THEORY OF WAVE MOTION 

distance of the point M from its mean position is called the 
amplitude. 

101. The period and phase constant. When the angle 6 
has increased from its initial value a by 2 it radians, .the 
point P will have described a complete circumference and 
the motion of the point N will have passed through all of 
its phases. The time T, which is required to accomplish 
this, is called the period of the simple harmonic motion. The 
period is determined by the condition that the angle co T de- 
scribed by the point P in the time T must be equal to -2 it. 
Therefore we find 

(1) coT=2ir, T=—. 

CO 

If we wish to put the period into evidence in the equation 
of a simple harmonic motion, we observe that (1) gives 

2tt 
T 

so that we may write, in place of (3), Art. 100, 

(2) y = asin(-^+a 

This equation represents a simple harmonic motion of 
period T and of amplitude a. The quantity a is called the 
phase constant. The phase constant is an angle and enables 
us to calculate the distance from to the position occupied by 
the moving point iVat the time t = when the motion began. 

Thus, if a = 0, the point N starts from as its initial position and 
begins to move upward ; if a = -, the initial position of N is at B, etc. 

Let us think of two different points oscillating up and 
down along the line BB\ each in accordance with an equation 
of the form (2), the amplitudes and periods of the two mo- 
tions being the same while the phase constants are different. 
Then these two points will move in exactly the same way, 
only that one will always be ahead of the other. The sec- 
ond point will appear to be imitating the motion of the first, 



ILLUSTRATIONS OF HARMONIC MOTION 229 

lagging behind it in a perfectly definite fashion. This is 

what is meant by saying that the phases i){ the two motions 

are different, the phase oi the motion (2) at the time t being 

equal to , 

2 irt . 

-Y + n - 

The use of the word phase in this connection is not merely accidental. 
The appearance of the moon at a given instant (its phase) depends 
upon the place in its orbit around the earth which it happens to occupy 
at that time. By analogy we speak of a periodic phenomenon as passing 
through all of its phases in the course of a period, and of course two 
otherwise identical periodic phenomena may have their corresponding 
phases occur at different times. It is this difference which manifests 
itself in the different values of the phase constant. 

We have already observed that the substitution « = — 4- a' 

& >nverts (3), Art. 100, into an equation of form (2), Art. 100. 
We may now express this fact as follows: 

Tlie two equations 

. -2 irt 2 irt . f2wt , ir 

represent two simple harmonic motions of amplitude a and 
1 1\ which differ only in phase, the phase difference being 

equal to — radians or 90°. 

The time interval which elapses between corresponding 
phases of these two motions is J T, that is, a quarter period. 

102. Some illustrations of simple harmonic motion. The 
notion of simple harmonic motion is of fundamental impor- 
tance in many problems of applied mathematics. The mo- 
tion of a simple pendulum, the vibrations of a tuning fork, 
and many of the motions of elastic bodies may be described 
conveniently in terms of simple harmonic motion. The 
vertical motion of a particle of a water wave is approximately 
of tin- same type, and the whole theory of sound and light is 

9ed on the idea of harmonic motion. 



230 THEORY OF WAVE MOTION 



EXERCISE LVIK 

In Exs. 1-5 the unit of time is one second and the unit of length 
one inch. Describe completely the simple harmonic motion given by 
each of these equations; that is, determine their amplitudes, periods, 
and phase constants, assuming equation (2) of Art. 101 as the standard 
form for the equation of such a motion. 

1. y = 2.3745 sin ^. 
y 5 

2. y = 3.7216 sin Ut + ^ 

3. y = 1.4712 sin (2.7215 1 + 1.7291). 

4. y = 11.7261 cos (7.0- 

5. y = 2.7268 sin f 2 vt + 13°) . 
J \ 16.7214 / 

6. A certain pendulum has a period of oscillation of 5 seconds. Its 
motion is started by displacing the bob from its position of equilibrium 
three inches toward the right and then releasing it. Write the equation 
of its motion. What will be the corresponding equation for a point 
halfway between the lower end of the pendulum and its point of 
support ? 

Hint. Since the amplitude of the oscillation is small as compared 
with the length of the pendulum, the motion may be regarded as taking 
place approximately in a straight line. Take this line as z-axis, positive 
toward the right, and choose as origin the point of equilibrium. Let the 
time t be measured in seconds from the moment in which the pendulum 
is released. Then the required equation for the lower end of the pendu- 
lum will be 

.-.-.(¥+!)■ 

7. A cork is bobbing up and down owing to the passing water waves. 
These waves are 4 inches high {i.e. the difference of level between the 
crest and trough is 4 inches). If seventy of these waves pass in a 
minute, and we start to count time from one of the instants when the 
cork has reached its highest position, what equation will describe the 
motion of the cork approximately ? 

8. Show that the following mechanism enables us to convert uniform 
circular rnotion into simple harmonic motion. RR' (Fig. 128) is a rod 
which may slide back and forth in its own direction, its motion being 
limited by the guides A, B, A', B' . Attached to the rod there is a slot S 
perpendicular to the rod. A crank C, moving with uniform velocity in a 



SiMPLK HARMONIC CURVES 



231 



circle around the point (\ is made to tit accurately into the slot S. Every 
point of the rod HI! will then describe a simple harmonic motion. 







>' 












-„ 




o 




-v 


s s 








v \ 


\ 




B 

ZZ1 






^- A- 


n 


K 






\ ^ 




zn 


\ 




1 


□ 


.1 


•- 




/ 


A 



Fig. 128 

9. Show that a point on the piston rod of a steam engine will move to 
and fro approximately according to the law of simple harmonic motion. 

103. Simple harmonic curves. In the equation charac- 
teristic of a simple harmonic motion, namely, 

y = a sin {cot + «), 

let as substitute x in place of t and interpret x and?/ as the 
rectangular coordinates of a point in a plane. The curves 
obtained as a result of plotting such equations, 

y = a sin(axr + a), 

are called simple harmonic curves. 

We may also think of the relation between simple harmonic motion 
and simple harmonic curves in the following more concrete fashion. At- 
tach a light pin P (Fig. 129) to one of the 
prongs of a vibrating tuning fork, and allow 
it to press lightly against a strip of smoked 
If this strip of glass is at rest, the pin 
will make a Bhort straight line upon it. But 
if the strip be moved with a constant velocity 
in a direction perpendicular to that of the 
vibration of the tuning fork, the point P 

will describe a wavelike curve upon the slide. This curve may be 
record of the motion of the tuning fork. It is easy fo see 
that the record produced in this way by any simple harmonic motion is a 
simple harmonic curve 

The simplest case of a simple harmonic curve is that of 
the sine curve = sm ^ 




Fig. 129 



232 



THEORY OF WAVE MOTION 



whose form has, by this time, become familiar to the student. 
(Compare the middle curve in Fig. 130.) It has the form 
of a wave line with nodes at the points x — 0, x = ± it, 
x= ± 2 7r, etc., with crests or maxima one unit high above the 

P° ints _tt _5,r _9,r . 

x — — , x — — — , x — — — , etc., 

of the #-axis, and with troughs or minima one unit deep 
below the points 



x = 



IT 



X = 



t IT 



11 



etc., 




Fig. 130 



2 2 2 

The length of one complete undulation of the 
curve is equal to 2 it. 

104. Amplitude. It is clear 
that the curve 

(1) y = a sin #, 

c where a is any fixed positive 
number, is of the same general 
form as the sine curve. It 
has the same nodes (points of 
intersection with the #-axis), 
and each of its undulations has the same length as that of 
the sine curve. Its maxima are above the same points of the 
#-axis as those of the sine curve, but they are higher or lower 
according as a is greater or less than unity, a is called the 
amplitude of the curve. Figure 130 shows three such curves 
of amplitude, \, 1, and 2. It is clear, then, that the depth of 
the wavelike curve (1) is dependent upon the value of the 
amplitude. 

105. Wave length. The two curves 

y = sin x and y = sin 2 x 
are shown together in Fig. 131. The latter curve has the 
same general form as the former, but each of its undulations 
(its wave length) is only half as long. 

Similarly, the curve 
(1) y = sin nx, 



PHASE CONSTANT 



233 



where n is a positive integer, is found to be a wave line of 
the same height as the sine eurve, but having w eomplete 
undulations between a? = 
and x=2ir. Consequently 
its wave length X is given by 

(2) X = fZ. 

Equation (1) represents a 

simple harmonic eurve of 




wave-lenerth X 



, even if 



Fig. 181 



n is not an integer. For the function sin nx will pass through 
all of its values just once, while nx changes from to 2 it ; 

that is, while x changes from to 



'ITT 



We may put the wave length into evidence in the equation 
<>f the curve, by solving (2) for n and substituting the result- 



ing value of n in (1). We find n 
(••)) y = sin 



, and therefore 



7r.r 



as the equation of a simple harmonic curve of wave length X. 
If we combine this result with that of Art. 104, we see that 



O) 



y = a sin 



represent% a simple harmonic carve of ivave length X and of 
ampUtud* 

106. Phase constant. If we compare the curves 

y = sin x and y = cos x, 

we observe that they arc identical in form. That is, both 
are simple harmonic curves of the same wave length and 
amplitude. They differ only in position. We can, in fact, 
slide one of these curves along the z-axis in such a way as t<> 



234 



THEORY OF WAVE MOTION 



make it coincide with the other. This, as we have observed 
before (Art. 78), is the geometrical significance of the 
equation 



[Z + % 



COS X. 



We may therefore dispense with the cosine curve altogether 
and regard it as a displaced sine curve. More generally, the 
same thing is true of the curve 

(1) y=±= sin (*+«)> 

which coincides with the sine curve if a = and with the 
cosine curve if a=—. Equation (1) represents a sine curve 

Li 

displaced toward the left through a distance of a units. The 
quantity a is called the phase constant of this curve. 

If we combine the results of Arts. 104 and 105 with our 
latest remark, we see that 



(2) 



,1TTX . 

y = a sin ( — — -h a 



represents a simple harmonic curve of wave length X, amplitude 
a, and phase constant a. 

This equation may be put into a different form by making 
use of the addition formula of the sine. (See Art. 79.) For 
we have 



2 irx 



. z irx 
y =a\ sin cos a -f- cos 

A, A. 



sm a 



Consequently, if we put 

(3) m= a cos a, n = a sin a, 

we shall find 

,. N . 2 irx . 2ttx 

(4) y = m sin 1- n cos 



Conversely, any equation of the form (4) has a simple har- 
monic curve as its graph. 



WAVE MOTION 235 

For if m and n are given, we may compute a and a from 
(3), thus obtaining the value of amplitude and phase con- 
stant. Since a is positive, we have, from (3), 

C ") a = + \ //r + n a , tan a = — 

the quadrant of « being determined from the sign of its sine 
and cosine as given by (3 >. 

EXERCISE LVIII 

Draw the simple harmonic curves which correspond to the equations 
given in Exs. 1 to 5 : 

1. y = 2 sin U x + ?V 3. y = 3.1 sin (7.6 .r + 6.2). 

2. y = 3 cos (2 ,- - |). 4. y = 2.8 sin (^ + 72°). 

5. // = sin (./• -f- ~). y = -■ sin .r. 

6. In the general theory of simple harmonic curves we assumed a to 
he a positive quantity. Show that this assumption does not, after all, 
really exclude from consideration those cases in which a is negative. In 
other words show that a simple harmonic curve, for which a is negative, 
coincides with another one for which a is positive and whose phase con- 
>tant differs from that of the first curve by it radians. 

7. Write the equations of the curves of Exs. 1 to 5 in the form (4) of 
Art. 106. 

8. A simple harmonic curve is given by the equation 

y = 2.75 sin ^ - 3.76 cos 2 -^. 
5.76 •>..<'> 

mine its wave Length, amplitude, and phase constant. 

9. Discuss in the same way the equation 

y = 0.72 cos (7.52 ./•)— 2.07 sin (7.52 x). 

107. Wave motion. Our use of the word wave, in con- 
tion with simple harmonic curves, is not (juite in accord- 
ance with tin- accepted meaning of this term. Ordinarily 
when we speak of a wave, a water wave, for example, we 
mean a peculiar kind of motion. If the cross section of the 
surface of the water at a given instant is a simple harmonic 



236 THEORY OF WAVE MOTION 

curve, we should properly speak of this curve, not as the 
wave, but as the instantaneous profile of the wave.* It is 
characteristic of a wave that this profile is in motion. 

We shall obtain an excellent idea of wave motion by 
allowing a simple harmonic curve to glide along the rr-axis 
with a uniform velocity v. We shall call such a wave a 
simple harmonic wave, and v its velocity of propagation. 

Let t denote the time (expressed in seconds), and let us 
assume that v, the velocity of propagation (expressed in feet 
per second), is positive, so that the wave advances in the 
direction of the positive a>axis. Let the simple harmonic 
curve 

(1) y = asm(—- + a) 

be the wave profile at the time t. The profile of the wave 
at the time t=0(t seconds earlier) was a curve of the same 
form as (1), but situated farther toward the left. Therefore 
its equation can differ from (1) only in the value of the 
phase constant. Consequently we may assume that 



(2) y = a ami-—— + a 

is the equation of the wave profile at the time t = 0. We 
wish to find the relation between a, a , v, X, and t. 

The nodes of the wave profile at the time t = 0, that is, its 
intersections with the #-axis, are obtained from (2) by 
equating y to zero. These nodes (see Fig. 132), the points 
M , M ', M ", etc., are infinite in number, and the distance 
between two consecutive ones is equal to \ X or one half of 
the wave length. One of these nodes, M say, will be ob- 
tained by equating (- a to zero. Since OM is the 

X 

abscissa of the point M and since for this point 
2 irx n 



* The wave profile is the cross section which one would obtain of the surface 
of the water if it were to freeze suddenly while a wave is passing. 



WAVE MOTION 



237 



we shall have 
(3) 



03L 



*-«o 



After f seconds, the wave profile has moved from its original 
position M 4 V, ••• fco MAM 1 •••. The equation of the wave 
profile is now given by (1). 
The nodes of this profile 
are the points of the curve 
for which 



77. r 



■f a = fr 




where & is either equal to 

zero or to a positive or 

negative integer. Consequently these nodes are the points 

of the 2-axis whose abscissas have the values 

x = - £L + I frX, where h = 0, ± 1, ± 2, ± 3, .... 

- 77 

One of these points is the new position M occupied by the 
point M Q as a consequence of the motion of the wave profile 
from MlAJMJ ... to MAM 1 •••. Since OM is the abscissa of 
31. we shall therefore have 



(4) 



2 7T 



where h is a definite positive or negative integer or zero 
whose precise value remains to be determined. 

But if v is the velocity of propagation of the wave in feet 
per second, every point on it moves through a distance of vt 
feet in t seconds. Therefore 






031- 03L = vt. 



If we substitute in this equation the values (3) and (4) for 
031 

(6) 



031 -and 03I . we find 



-±ZL+lk\=vt 



-7T ITT 



238 THEORY OF WAVE MOTION 

This equation must be true for all values of t. In particular 
it must be true for t = 0. But for t = we have a = « , so 
that the equation involves a contradiction unless k = 0. 
Consequently the integer Jc which appears in equations (4) 
and (6) must be equal to zero, and we have 



whence 
CO 



Xa n Xa , X , N . 



2irvt 
a = a A 



If we substitute this value of a in (1), we find 

"2tt 



(8) y = a sin 



— O - irt) + a 



as the general equation of a simple harmonic wave of amplitude 
a and wave length X, whose velocity of propagation is equal to v. 
We may still speak of a as the phase constant. It is the 
phase constant of the wave profile at the time t = 0. The 
phase constant of the profile curve at any other instant may 
be computed from (7). 

If in (8) we assign a fixed value to t, we obtain the equa- 
tion of the wave profile at that instant. Let us instead 
assign a fixed value to #, so that y becomes a function of t 
alone. In the case of a water wave this would amount to 
a study of the upward and downward oscillations of a cork 
floating upon the water. We shall naturally inquire as to 
the length of time which is required to complete such an 
oscillation. This time is called the period and may be 
denoted by T. Clearly T is the time which must be added 
to t so as to change the argument of the sine function in (8) 
by ± 2 7r. Now if we increase t by T without changing rr, 

this argument changes by — , and this will be equal to 

-27rif X 

— = 1, or T= -. 

X v 



GENERAL HARMONIC MOTION 239 

Therefore, if r is the velocity of propagation, X the wave 
lengthy and T the period, we have the relation 



{9) X=Tv. 

Instead of (8) we may now write 



(10) y = a sin 



"a 



j-iH 



as the general equation of a /rare of length \, period T, ampli- 
tude (?, and phase constant a . 

We Bee finally that a simple harmonic ware has for its profile 
at any moment a simple harmonic curve, and that every point 
upon it oscillates up and down in accordance with the law of 
simple harmonic motion. 

EXERCISE LIX 

If the units of length and time are feet and seconds respectively, com- 
pute the wave length, period, amplitude, and phase constant of each of the 
waves represented by the following equations : 

i !-•-.[■ -(§-£)+=]. 

2. y = 2siD fo*-7l+|Y 

3. y = a sin (Jix + ct + d). 

108. General harmonic motion. A simple harmonic motion 
is the simplest case of an oscillatory phenomenon, and many 
natural periodic motions may be adequately described as 
simple harmonic motions. We have already given some ex- 
amples of such cases. But very frequently the • motion, 
although of an oscillatory character, is more complicated. 
In tin- case of a water wave, for instance, we see that the 
profile of the wave is not a simple harmonic curve, but that 
there are smaller waves (ripples so to speak) running along 
the backs of the Larger ones, thus complicating the motion. 
Simple experiments show that the sound waves produced by 
a tuning fork are very approximately represented by simple 
harmonic motion; but other musical instruments, such as the 



240 THEORY OF WAVE MOTION 

violin, the piano, the human voice, produce sound waves 
which resemble the more complicated water waves. 

A tuning fork which makes 129 oscillations in a second 
causes a certain simple harmonic motion of the air particles 
whose period is y^gth of a second and which produces a 
certain tone usually denoted b} T C. If the same note is 
struck on the piano, it is found that the principal part of the 
motion of the air particles again has yjgth of a second as 
its period. But the motion is not simply harmonic. It is 
a combination of this fundamental motion with one twice as 
fast, with another three times as fast, and so on. In other 
words, the motion of the air particles is given by an equation 
of the form 

(1) y = a x sin C-^t + aA + a 2 sin (-^t + a. 



a* sin — — 

3 ^ rp 



"■* + «.]- 



where the period of the first and principal term is T, that of 
the second J T, that of the third | T, etc. 

This is not the place to discuss details of the theory of 
sound. Our purpose in entering upon this theory at all 
was merely to explain one of the many instances in which 
sums of simple harmonic functions of the form (1) present 
themselves as indispensable. 

We wish to learn how the various .terms in (1) combine. 
For that purpose the length of the period T makes but little 
difference. We shall therefore put 

T=2tt 
since the formulas will then assume a somewhat simpler 
appearance. Then (1) reduces to 
(2) y = a x sin (t 4- a x ) + a 2 sin (2 t + a 2 ) 

+ a 3 sin (3 t -f- a 3 ) + ••• • 
Now each of these terms may be expanded in accordance 
with the addition theorem (Art. 79), so that 
a x sin (t -h a 1 ) = a x cos a x sin t + a 1 sin a x cos £, 
a 2 sin (2* + a 2 ) = a 2 cos a 2 sin 2 t 4- a 2 sin a 2 cos 2 t, etc. 



GENERAL HARMONIC CURVES 241 

Consequently, if we introduce new constants A v A T ••• , B v 
B r ••• by putting 

A x = a l sin a v A 2 = a 2 sin o^ ••• , 
B 1 = (7j cos a v B 2 = a 2 cos ^ ••• , 
equation (2) becomes 

( .,. y = A l cos f 4- A 2 cos 2 f H- ./i 3 cos 3 t+ ••• 
4- B x sin « 4- i? 2 sin 2 ^ 4- i? 3 sin 3 t -\ . 

Let us call the fixed point, with which the moving point 
would tend to coincide if the amplitudes a v a 2 , etc., of all the 
simple harmonic motions of (1) were to approach the limit 
zero, the center of oscillation. We have tacitly assumed so 
far that the center of oscillation was the origin of coordinates. 
Let us drop this specialization, and let \ r A be the fixed 
value to which y would reduce if all of the oscillations were 
to disappear : that is, let J A be the ordinate of the center 
of oscillation.* Then we must add J A to the right member 
of (3), so that we obtain finally 

(i\ y = \ ^o + A cos f + ^2 cos 2 £ + ^4 3 cos 3 H 

+ B x sin t + B 2 sin 2 t + B 3 sin 3 t -\ 

as the typical equation of a general harmonic motion. 

A- in the case of a simple harmonic motion, we may make a graphical 
record of this motion. Suppose, for instance, that the motion to be in- 
gated is the vibration of a metallic wire. We attach a light pen to 
the wire so as to enable it to write upon a strip of smoked glass. If the 
be left at rest while the wire is caused to vibrate, the pen will 
merely describe a straight line upon the smoked glass. If, however, the 
_ sa be moved with a rapid uniform motion at right angles to the 
direction of vibration of the wire, there w T ill appear as record a wavelike 
curve. This curve will belong to the class considered in the next article. 

109. General harmonic curves. If we put x in place of t 

in equation (4) of Art. 108, we find 

, -. n .'/=•] A () 4- A x cos x 4- A 2 cos 2 x 4- A s cos 3 2; + 

4- B x sin x 4- B 2 sin 2 x + B 3 sin 3 x 4- ••• • 

* The reason for denoting this quantity by \ A rather than A will appear 
later. (See Art : 



242 



THEORY OF WAVE MOTION 



The curves which are obtained as a result of plotting an 
equation of this form are called general harmonic curves and 
are capable of an extraordinary variety of forms. In fact it 
can be shown, by methods involving the integral calculus, 
that an infinite series of the form (1) may be found to repre- 
sent almost any continuous curve, and even extensive classes 
of discontinuous curves *. In this book, however, we are con- 
cerned only with sums of the form (1) involving a finite 
number of terms and the curves represented by them. The 
name harmonic curves will be understood to apply only to 
such curves. 

We proceed to discuss an example. Let us plot the curve whose 
equation is 

(2) y = sin x + sin 2 x. 
We begin by drawing the two familiar curves 

(3) y x = sin x and y 2 = sin 2 x, 

the two dotted curves of Fig. 133. From these curves it is easy to con- 
struct the curve (2). For we see from (2) and (3) that 

V = Vi + 1/2 
for every value of x. If then we find the ordinate of each of ihe two 
dotted curves for a given value of x, their algebraic sum will be the 
ordinate of a point on the required curve. The resulting curve is in- 
dicated in Fig. 133 by a full line. A few points of this curve may easily 




Fig. 133 



* Such series are usually called Fourier's series in honor of the great mathe- 
matical physicist who first stated, and in part proved, the above theorem. The 
first rigorous proof was furnished much later by Dirichlet. 



TRIGONOMETRIC INTERPOLATION 243 

be obtained by inspect ion. For x = 0, y, and //., are both zero, and there- 
fore also v = t/ l + >/., — 0. Consequently the point f> is on the curve. 
For j- = - 2, jfj = 1 and y s = 0, so that y = 1 and the curve passes through 
the point .1. For values of x between tt 2 and it, y a is negative so that 
f a will be less than y r Consequently the full line curve in this in- 
terval lies below the corresponding portion .!/> of the curve i/ l — sin x. 
For a certain value of X in this interval (determined by the equation 
sin x + sin "J r — <>) y, and //., will be numerically equal but opposite in 
rign, so that at that point the curve y = sin x + sin 2 x will cross the 
i-axis. This is the point L of Fig. 133. For any value of x we may 
obtain y, and y 9 by measurement from the two dotted curves. If we 
form the sum of these two quantities with due regard to sign, we find 
the corresponding ordinate of the required curve.* 



EXERCISE LX 
Plot the following harmonic curves : 

1. y = 2 sin x -f sin 2 x. 3. y = sin x + cos 2 x. 

2. y = sin z + i sin 2 ./•. 4. y = 5 sin x 4- sin 4 x. 

110. Harmonic analysis or trigonometric interpolation. We 
have seen how a number of simple harmonic curves may be 
compounded into a single general harmonic curve. It often 
happens that a curve is given, actually drawn out on paper, 
as for instance in the case of a self-recording barometer or 
thermometer. If the curve is of a periodic character, the 
question arises whether it may be regarded as a harmonic 
curve, that is, whether it is possible to compound it out of a 
number of simple harmonic curves by the method of Art. 
109. And if so, the problem presents itself to actually find 
tin' component simple harmonic curves. The process of 
solving this problem is known as harmonic analysis and is of 
!t importance in many branches of pure and applied 
mathemal 

Let ii- suppose that the given curve is periodic, so that it 



.ii'l Sti: \tton have devised a machine for performing mechan- 
ically the operation «>f comhining a number «>f rimple harmonic curves. This 
niachin-- is also capable "f performingthe inverse operation discussed in Art. 110. 
For this reason it baa been called a harmonic """ 



244 



THEORY OF AVAVE MOTION 



consists of an infinite number of equal pieces, and let the 
length of one of these pieces, the wave length, be equal to X. 
If \ is less than 2 7r, we may, by a process of magnification 
or stretching, replace the given curve by another one similar 
to it whose wave length is just equal to 2 it. If we can solve 
our problem for this second curve of wave length 2 7r, it will 
be easy to solve the corresponding problem for the original 
curve. If A, is greater than 2 7r, a process of compression 
enables us to reduce the problem again to the case of a curve 
of wave length 2 ir. We may therefore assume that the 
wave length of the given periodic curve is equal to 2 it 
without reducing, by this assumption, in any essential 
fashion the general applicability of our results. The ob- 
ject of this assumption is merely to simplify the resulting 
formulse. 

Let P P 1 P 2 ••• (Fig. 134) be the given curve of wave 
length OA = 2 7r, and let us divide OA into an odd number, 




+x 



Fig. 134 



say 2 m + 1, of equal parts. Not counting A, there will then 
be 2m + l points of division; namely, 0, X v Jf 2 , ••• X 2m . 
In Fig. 134 we have made 2 m + 1 = 7. 

At these points of division we construct the ordinates 

OP , X t P v X 2 P 2 , ••• 

of the curve. Let y , y v y 2 , ••• y 2m be these ordinates, each 
with its proper sign prefixed. On account of the periodic 
character of the curve, the ordinate at A will be the same as 
that at 0. This is the reason that we did not count A as 



TRIGONOMETRIC ENTERPOLATION 245 

one of the points of division. If we had included A, we 
should really have been counting twice. 

We ran always find a harmonic curve involving terms in 
x. - .r, 3 .i\ --• »u\ which passes through the 2m + l points 
P , P v P~ ••• P 2m . For the general equation of such a 
harmonic curve is 

,,n y = \ A + <4.j cos x + Jl 2 cos 2 x + ••• + A„ (,()S »'•>' 
+ B x sin x + i? 2 sin 2 x + ••• -h ^ w sin wa?, 

and therefore contains 2w + l coefficients ^4 , tIj, •••-4 m , 
2? r ••• i? m . which may be determined in such a way as to make 
the corresponding curve (1) pass through the 2 m + 1 given 
points. In fact, the curve (1) will pass through the point 
P , if the value of y obtained from (1), for x = 0, is equal to 
the ordinate y of the given point P ; that is, if 

(2) y =£4> +A + 4j+ ••• +4». 

The abscissa of P t is OX, = — -. Therefore the curve 

1 ' 2 m + 1 

(1) will pass through P r if the value of y obtained from 
(1 ). for .?• = w — , is equal to the ordinate y 1 of the given 
point ]\ ; that is, if 

(3) fc - Mi+ ^„ i li I+ j,«.£JLi 

+ ••• + A*cos- — 

2 7W + 1 

•2 • 2 7T 



+ ^1 sin — — + B 2 sin ~ ~ 

1 2 w -f- 1 2 2 w + 1 



+ ... +P m sin 



9 



2m-M 



In the sain.- way we find that the curve (1) will pass through 
the points P 2 . P v ... P 2;n if the following additional equations 
are satisfied : 



246 



THEORY OF WAVE MOTION 



(4) 



y 2 = 1 A + A 1 cos 



2ra-f 1 



A cos + 

2 2m + l 



,. W • 4 7T , -r> • 47T , D 2 • 4 7T 

+ A» cos ^ — — — + -»i sin o 7 "+" ^2 Sln 



2m + l 



2m+l 

+ ••• +2? m sin 



2ra + l 

m • 47T 

2m + l' 



i a , a 2 m7r , A 2*2 m7r 



+ A m cos 



2m + 



~t + A sin q — r^r + ^2 S111 « — — r 

1 2m+l 2m+l 

m • 2m7r 
2w+ 1 



m + 1 

H hi? TO sin 



Now the equations (2), (3), and (4) are together 2 m + 1 in 
number, and involve 2m +1 unknown quantities, viz., the 
quantities, A , A v ••• J. m , 2? r ••• B m . The coefficients of these 
quantities (sines and cosines of known angles) are known 
numbers, and the quantities y , y v ••• y 2m , the ordinates of 
the given points P , P v • •• P 2m , are known. We may there- 
fore, in general, solve these 2m + l equations for the 2 m + 1 
unknown quantities and then substitute the values obtained 
for A , A v ••• A m B v ••• B m in (1). The harmonic curve 
obtained by plotting (1) will then pass through the given 
points. 

If the given curve is fairly smooth, and if the points P , 
P v ••• P 2m be taken close enough together, that is, if m be 
chosen large enough, we may find by the above method a 
harmonic curve which may be regarded as replacing the 
original curve everywhere with any desired degree of ap- 
proximation.* 

The following example will illustrate this method. Let y be a peri- 
odic function of x, of period 2 ir, such that the values of y which corre- 
spond to 

x = 0°, 72°, 144°, 216°, 288° 



* A more precise formulation of this statement must be left for a much later 
stage of the student's mathematical career. 



TRIGONOMETRIC [NTERPOLATION 247 

arc respectively 

// = (), + 2. +i - |, -2. 

Our general theory tells us that wo may find an expression of the form 

// = i .1 + A i cos x + A- cos 2 x + 7>i sin x + Ik sin 2 x } 

which assumes the live \ alues assigned to y for the five given values of x. 
Moreover we have the following five equations for the five unknown co- 
efficients .1 0< .1,. .1... B h 1U : 

2 = i An ■ . 1 1 eos ?•_> + . 1 , cos 144° + £i sin 72° + Bi sin 144°, 
| = \ Aq + A\ cos 144° + .1-2 cos 288° + P n sin L44 + B 2 sin 288°, 

-| = 1 ;t + -li cos 'JIG- + .1-2 cos 72° + Bi sin 216° + £ 2 sin 72 . 

- 2 = \A Q + A i cos 288 . 1 i eos JIG 5 + />', sin J8S l> + £ s sin 216°. 

Since we have 

sin JIG = sin (360° - 144°)= - sin 144°, 
cos 216° = cos (360° - 144°) = cos 144°, 
sin JSS = sin (360° - 72°) = - sin 7J°, 
cos 288° = cos (360° - 7J°) = cos 72°, 

the above equations may also be written as follows : 

(1) \Ao + A x + A % = % 

(J) |i + A 1 eos72° + A % cos 144 + B x sin 72° + B 2 sin 144° =2, 

(:)) \A„ - .Ij cos 144° + -4, cos 7 J 3 + B x sin 144° - B 2 sin 72° = a, 
(4) \A* + .4! cos 144° + 4« cos 72° - 7?! sin 144° + 7i 2 sin 72° = - *, 
I o + - 1 1 cos 7J 3 -f Ai cos 144° - B\ sin 72° - £ 2 sin 144° = - 2. 

From (J) and (.">) we find by addition 

\ Ao 4- -li cos 72° + .4 2 cos 144° = 0, 

and similarly from (8) and (4). 

(7) | Jo + Jicos 144 D + yl 2 cos7J° = 0. 

From (1 ) we have 

| 4 = - A\ - A 2 

which, substituted in (G) and (.">), gives 

(8) .l,(ros7J~ - 1)- .1 2 (cos 144"' - 1) =0, 
.li (coe 144 - 1)+ .1 2 (cos 72°- l)-0. 

If we multiply both members of the first of these equations by cos 72° — 1, 
those of the second by - (cos 111 1 ). and add. we find 

(•) ,1 ,[(<■<,. 72 - l) 2 -(cosl44 3 - l) r \ = 0. 

From the table of natural functions, we find to two decimal places 

cos 72° = 0.31, cos 144° = - cos 36° = - 0.81. 



248 THEORY OF WAVE MOTION 

Therefore 

cos 72° - 1 = - 0.69, cos 144° - 1 = - 1.81, 

so that the coefficient of A x in (9) is not equal to zero. Consequently 
we conclude from (9) that A 1 = 0. According to (8) and (1), we must 
then have also A 2 — 0, A — 0. 

If now we put A = A x = A 2 = 0, in (1) to (5), these five equations 
reduce to the following two : 

B, sin 72° + B„ sin 144° = 2, 
(10) x 2 

v . B 1 sin 144° - B 2 sin 72° = \. 

From these equations we eliminate first B 2 and then B 1 , giving 

(sin 2 72° + sin 2 144°)5 1 = 2 sin 72° + \ sin 144°, 

( ^ (sin 2 72° + sin 2 144°) B 2 = 2 sin 144° - \ sin 72°. 

From the table of natural sines we find, correct to two decimal places, 

sin 72° = 0.95, sin 144° = sin 36° = 0.59, 
so that 

sin 2 72° + sin 2 144° = 0.90 + 0.35 = 1.25. 

Consequently, equations (11) become 

1.25 B 1 = 1.90 + 0.30 = 2.20, 

1.25^ = 1.18 -0.48 = 0.70; 
whence finally 

(12) B 1= 1.76, £ 2 = 0.56. 

Since we have already found .4 = A x — A 2 = 0, the function which we 
were seeking is 

(13) y = 1.76 sin x + 0.56 sin 2 x. 

In order to check our result we may substitute the five given values of 
x in (13) and verify that the corresponding values of y are actually those 
which were originally given. That equation (13) gives y = for x = is 
obvious. For x — 72° and for x = 144°, we find from (13) 

y = 1.76 x 0.95 + 0.56 x 0.59 = 1.67 + 0.33 = 2.00, 
and 

y = 1.76 x 0.59 + 0.56 x (- 0.95) = 1.04 - 0.53 = 0.51, 

respectively, checking to within one unit of the last decimal place 
computed. To check the other two pairs of values requires no additional 
computation. 

In this example, the values of y for x = 144° = 180° - 36° 
and for x— 216° = 180° + 36° were numerically equal but 
opposite in sign. The values of y for x = 72° = 180° - 108° 
and for x = 288° = 180° +■ 108° were also numerically equal 



TRIGONOMETRIC INTERPOLATION" , 249 

but opposite in sign. It is owing' to this circumstance that 
Am A v and A 2 turned out to be all three equal to zero. 
Having- become aware of this fact, we may abbreviate our 
work very much by at once equating A , A v A v etc., to zero, 
whenever the given values of the function are numerically 
equal but opposite in sign for those among the given angles 
which differ numerically by the same amount from 180°. 

Similarly, if the given values of the function y are equal 
numerically and in sign for all of those among the given 
angles x which differ numerically by the same amount from 
180°. the expression for y will contain no sine terms ; that is, 
B v B v B z , etc., will all be equal to zero. 

EXERCISE LXI 

1. Find a periodic function, involving only the angles x and 2 x, 
which assumes the values 

y = 0, + 2, .+ 1, - 1, - 2, 

for x = 0°, 72°, 1U°, 216°, 288°, 

respectively. Compute the coefficients to two decimal places. 

2. Find a periodic function, involving only the angles x and 2 x, 
which assumes the values 

y = + 2, + 1, - h - h + 1, 
for x = 0\ 72°, 144°, 216°, 288°, 

sctively. Compute the coefficients to two decimal places. 

111. Theorems leading to the general solution of the problem 
of trigonometric interpolation. We have shown in Art. 110 
that the problem of trigonometric interpolation may be re- 
duced to that of solving a system of 2 m + 1 equations of the 
first degree with 2 m -f- 1 unknowns. But we can accomplish 
much more than this. We shall derive elegant and con- 
venient formula' for the solutions of these equations, en- 
abling us to find the values of the coefficients A k and B k by 
a direct and simple process. But before we can do this, we 
must prepare the way by proving some theorems necessary 
for this purpose. 



250 THEORY OF WAVE MOTION 

We begin by proving that the following formula 
(1) sin a + sin (a + f) -f sin (a + 2 t) + • • • + sin (a + mt) 
, / . mt\ . (w + l)t 



8in 2 
published by Euler * in 1743, is true for all values of a and £, 

provided that sin - is not equal to zero. 

-j * 

Proof. Let us denote the sum in the left-hand member 
of (1) by s m . If we multiply s m by 2 sin -, we shall have 

2 s m sin - = 2 sin a sin - + 2 sin (a + t) sin - 

Li A It 

+ 2 sin (a + 2 £) sin - + • • • +2 sin (a + ra£) sin - . 

Every term in the right member of this equation contains 
a product of two sines, and may therefore be expressed as a 
difference of two cosines by means of formula (4) of Art. 82 ; 
that is, 

sin a sin ft = \ [cos (a — /3) — cos (a -j- /3)]. 

We find, in this way, 

2 s m sin - = cos (a — %t)— cos (a + | t) 

+ cos (a + J — cos (a + § 
+ cos (a + | — cos(a'+ f £) 

+ 

+ cos \a + (m — ^)t\ — cos \a + (m + \)t\. 

* Euler (1707-1783) was born in Switzerland, but spent most of the years of 
his scientific career in St. Petersburg and Berlin. His work was of fundamental 
importance in all parts of pure and applied mathematics. Although absolutely 
blind during the latter part of his life, he continued to labor and to make im- 
portant contributions up to the end. 



TRIGONOMETRIC INTERPOLATION 251 

Clearly all of the terms in the right member except the 
first and last will destroy each other, so that we are left 
with the equation 

'2 s m sin - = co±{<< — .] O — cosja + (m + ) 2 )t\. 

But we have the formula (see Art. 82, equations (5)), 
D 7 . A-B A+ B 

cos A —cos B = — 1 sm COS : , 

so that 

2 s m sin - = — 2 sin - ( — t — mf) sin - (2 a + mi) 



= 2sm( « + — J sin * - — '-, 



whence, if sin - is not equal to zero, 

a + — J sm ^ — — ^~ 



sw = 



. t 

sin - 



This is the same as equation (1), the formula which we 
wished to prove. 

Let us rewrite formula (1) with a! in place of a, and then 
put 

«' = " + ?• 

Since sin a 1 = sin/ a + ^ ) = cos a, 

we then find 

(2) c m = cos a + cos (a + + cos (a + 2 f)H (-cos(a + mi) 

H sm ! — ^- 

27 2 



sin 

_ 



252 THEORY OF WAVE MOTION 

a formula which may also be obtained directly by a process 
strictly analogous to the one employed for the proof of 
equation (1). Formula (2} is also due to Euler. 
If in equations (1) and (2) we put a = 0, we find 

. mt . (m-\-Y)t 

sin — sin ^ — — — 

2 2 

(3) sin t+ sin 2 t+ sin 3 t-\ + sinm£ = 

sin- 

and 

mt . (m + V)t 
cos — sin ^ — y 

(4) l + cos£ + cos2tf + cos3£H |-cosra£ = ~ t ' 

sin - 

2 

Let us subtract \ from both members of (4). We obtain 
the formula 

mt . (m + 1)^ 
cos — sin ^ — — 

1 2 2 1 
- + cos£+cos2H h cosm£ = — - 

A t A 

sin — 

o mt . (m + l)t • t . / , 1\, , . < • t 

2 cos — sin — '- sin - sm [m + — )t + sm sin- 

2 2 2 V2/2 2 

2 sin - 2 sin — 

A 2 

where we have made use of one of the equations (4) of Art. 
82. The numerator of the last fraction obviously reduces to 
its first term, so that we find finally 

. (2 m + 1)* 
sm ^ - 1 — *- 

1 2 

(5) - + cos t + cos 2t-\- cos 3 t -\ — 4- cos mt = ; ) 

2 2 sin | 

A 

a formula which was known to Snellius in 1627. 

The angle t which occurs in all of these equations may 
have any value whatever excepting only those values for 

which sin - is equal to zero ; that is, excepting the values 

A 

2 for, where k is an integer. We shall now apply these formulae 



TRIGONOMETRIC INTERPOLATION 253 

to the particular case when t is a commensurable fractional 
pan of the entire circumference, so that 

t = — 1 IT, 

n 

where both //and n are positive integers and where 

n >1. 

We shall further put m = n — 1. 

We may express these assumptions more concretely as fol- 
lows. Let us divide the circumference of the circle into n 

equal parts, where n > 1. Then - — will be the angle sub- 

n 

tended by one of these parts. The smallest multiple of this 
angle which is equal to a complete circumference is of course 
the nth. Therefore the angles 

a n o o 

zJL, &JL, 3 -=^,...(n-l)2z 
n n n n 

are all distinct, and we propose to find the sum of their sines 
as well as of their cosines. We consider next the angle 

2 — ^ = — — and all of its multiples up to (w — 1) — — and calcu- 
n n n 

late the sum of their sines and of their cosines. In general, 

we consider the angle k- — = "" and its multiples 2 ~ , 

n n n 

3 "* ' lr — (n — 1)^— and compute the sum of their sines and 
n ?i 

of their cosines. 

According to ( ;1 > ) we have (putting t= - — and m = n— 1) 

2 kit . 2 far , . « 2 far , . . , -, - kir 

sin h sin 2 f- sin 3 1- ••• +sm(«-l) 

n n n n 

Cn — l)&7r . 7 
Bin - v z — sin kit 



■ hlT 

sm — 
n 

a formula which will be valid unless sin —is equal to zero; 

n 



254 THEORY OF WAVE MOTION 

that is, unless k is an integral multiple of n. Excluding 
this case for a moment, we see that the right member is equal 
to zero since sin kir occurs as a factor in the numerator and 
since the sine of any integral multiple of it is equal to zero. 
Consider now the excluded case when Jc is a multiple of n. 
Then every term on the left member is individually equal to 
zero. We see therefore that 

ra\ v • 2 hir . . 2 kir . . 2 klT . 

(6) 8 k — sin h sin 2 \- sin 3 h • • • 

n n n 

+ sin (n — 1) = 0, 

n 

for every value of k, whether k is divisible by n or not. 

We may prove in the same way, making use of equation 
(4), that 

/7 n n -, . 2 &7r . 2 kir , o 2 kir , 

(7) 6j.= l-fcos h cos 2 h cos 3 h ••• 

n n n 

, -j v 2 kir r, 

+ cos (n — 1) = 0, 

*n 

if k is not a multiple of n. 

If k is a multiple of w, all of the angles , 2- , etc., 

n n 

are integral multiples of 2 7r, so that each of the cosines which 

appears in C k is equal to unity. There are n terms in C k . 

Therefore 

(8) C k = 1 + cos h cos 2 h ... -f cos (^ — 1) =n, 

n n n 

if k is a multiple of n. 

Let us now consider two angles of the form and - — , 

n n 

and denote by C ki the sum 

. n . n ., . 2 kir 21tt , 2kir 21tt 

(9) (7^=1 + cos cos |-cos2 cos 2 

n n n n 

, -, . 2 kir , 1 .2?7r 

+ • • • + cos (n — 1 ) cos (n — 1 ) 

^ n 



TRIGONOMETRIC INTERPOLATION 



255 



A product of two cosines may be expressed as a sum by 
means of formula (3) of Art. 82 ; that is, 

cos a cos ft = .] [cos (a — ft) -+- cos (« 4- /?)]• • 

Therefore we may write, if we apply this formula to each 

term of C kl , 

C« = i[l+1] 

+ ,rreos - ( * -"' + «>» 2 c* + ^ l 



+ • 



cos 



g a(*-0« +0o . 2 *(*+for-| 



+ J 



COS 



, 1x2(6 -tar. , ...2(^ + 0^1 

(w — I) -^ l — h cos (n — 1) -^ — — ^— • 

n n J 



Now the sum of all of the terms in the first column may be 
equated to I C k _ t if we again make use of the notation C k de- 
fined by equations (7) and (8). Similarly we observe that the 
sum of the terms in the second column is ^ C k+l . Therefore 

(10) C kl =l(O i .. l +C k+l -). 
Consider now the expression, analogous to (7 W , 

x11 x a 2 kir . 2 lir . . 2 kir . 2 lir . 

(11) «y w = 8in sin (- sin 2 sin 2 f- ••• 

n n n n 



2 fa 



+ sin(n— 1) sin (w — 1) 

Since we have (see Art. 82) 

sin a sin ft = \ [cos (a — ft) — cos (a + /3)], 
we find 



for 



h-l = 2 COS 



■2(k-l)ir 



— cos 



2(k + l)ir 



n J 



+ o cos 2^ * cos 2 ! — f — 

L w n J 



4- |[c<» fa - 1) 2 ^-0^ _ cos ( n _ l 



2(fc + Z)7r 



256 



THEORY OF WAVE MOTION 



so that 

Finally let us put 
(13) (>S ft , Ci) = sin cos h sin 2 cos 2 h 



+ sin (n — 1) cos (n 



2_hr 
n 



Since we have (see Art. 82) 

sin a cos j3 = J [sin (a — /3) + sin (a -f- /3)], 

we find by a repetition of the above method 

(14) (&\,C l )=i(S h _ l +S h+l -). 

We have already shown (cf. equations (6), (7), and (8)) 
that 



(15) 



S k = for all values of &, 

C k = for all values of k which are not divisible by w, 

C k = n for all values of k which are divisible by n. 



If we make use of these facts, equations (10), (12), and 
(1-4) now teach us that the following statements are true. 



(16) 



(17) 



O kl = if neither k — I nor k -\- I is divisible by n, 

O kl = — if either k — I or k + 1 is divisible by w, but 

not both, 
C kl = n if both k —I and k + 1 are divisible by n. 

S kl = if neither k — I nor k + I is divisible by w, 

S kl = + — if k—l is divisible by n and k + I is not 
divisible by w, 

S H = — - if k—l is not divisible by n and k + 1 is 

divisible by n, 
S kl = if k — I and Jc + l are both divisible by n. 
(S k , C{) = for all values of k and I. 



TRIGONOMETRIC [NTERPOLATION 257 

112. General solution of the problem of trigonometric inter- 
polation. We are now ready tor the solution iA' the problem 
of trigonometric interpolation. We divide the circumference 
into - m 4- 1 equal parts by means of the angles 

fl "2 7T 4 7T () 7T 4 7717T 

' Im + V 2IH+1 1 2/w + l' '"' 2w + l' 
and denote the corresponding values of the function y by 

y ' yn fo // 3 * ••• ft. 

respectively. The function 

y = \ A + Aj eos x 4- -4 2 cos 2 x 4- • •• + vi„, cos waa; 
■f i>j sin x + i? 2 sin 2^4- ••• 4- 5 ffl sin wa; 

will actually assume these values, if the 2w+l quantities 
.4,,. A v • ••, J. m , J? x , i? 2 , • ••, B m are chosen so as to satisfy the 
'2 m + \ equations 

y = I A + A x 4- ^ 2 4- ••• 4- Am 

2 «7T ^ 7T 

y\ = l A o + *i cos /' 1 + ••• + ^ m cos m — - — 
2 m 4- 1 2 m + 1 



0> 



. » - 2w , D • 2tt 

4- B x sin — + ... 4- i? ro sin m 



W4-1 w> 2 m 4-1' 



1 1 i I 4 7W7T . . 4 ???7T 

#2m = § ^0 + A l C0S o TT + "• + A m C ° S m 



2m 4-1 2m + 1 

4- B x sm — — + ... + B m sin m - — — -, 

•1 in + 1 2 m 4- 1 

all of which may be expressed by the single equation 
(2) y 9 = J A Q 4- ^ cos 7 " 4- ••• 4- A m cos m 



w + 1 ' m 2w + l 

+ A 81D w- 7T + "' + 5 » sin m 75 TT' 

2m+l 2 m 4- ] 

if we think of i> as assuming in succession the 2 m 4- 1 values 

P=0, 1, 2, 3, -., 2 772. 

Let r be any integer between and m, let us multiply 

ii i 2 nr i 2 nr T 

//,, by 1. y l by cos — — , // 2 by cos 2 — — , ..., y v by 

m 4- 1 /// 4- i 



258 THEORY OF WAVE MOTION 

cos v -, ..., y„ m by cos 2 m ^ * ^ H , and add. If we 

m + 1 2 m 4- 1 

make use of equations (1) and the notations introduced in 

Art. Ill, we find 

( 3 ) Vo + y x cos ^ + ... + */ 2w cos2m 



2m + l ' **" 2m + 1 

= l^tf r + 4 c r lr + 4 2 <7 ar + ... +AAr 

From Art. Ill, (17), we know that all of the quantities 
(8 k , Ci) are equal to zero. Since r was chosen as an integer 
between and m, r can be divisible by 2 m + 1 (which num- 
ber corresponds to the n of equations (15), (16), (17) of 
Art. Ill), only if r = 0, In that case we shall have, accord- 
ing to (15) of Art. Ill, 











o r = 


-O = 


= ft = 


2m + 1, 




and 


accoi 


-ding 


to 


(16), 


Art. 


HI, 














^10 = 


= ^20 : 


= ... 


= W*o = 


0, 



since none of the numbers 1, 2, •••, m are divisible by 2m+l. 
Consequently, equation (3) reduces to 

(4) Wo + Vi + y a + - + ^ = J AC 2 ™ + 1) 

in the case r = 0. 

If r > 0, (7 r = 0. Moreover, O kr will be zero for all values 
of k for which neither k — r nor & + r is divisible by 2 m -J- 1. 
But & as well as r can at most be equal to m, so that neither 
k — r nor & + r is ever as large as 2m + l. The only case 
therefore in which one of these numbers can be divisible by 
2 m -f 1 is when k = r. Thus we have, according to (16), 
Art. Ill, 

C kr = for k different from r, 

n _2m + l 

0~- — r ♦ 



TRIGONOME rRIC DTTERPOLATION 259 

Consequently equation (3) gives, for r>0, 



2 ;*7r 2 m -f- 1 . 



and we notice that equation (4) may be thought of as 
included in (5) for /• = ().* We find therefore the formula 



(6) Ar = - 



m + 1 



nr 



2rir 



F. + Fi « , " + j + ft <** 2 2 ; + 1 + - 

o 2 r7r ~| 
+ F te ~am SSTT j 

(r = 0, 1, 2. .... m), 

enabling us to compute A , A l , • ••, A ra in terms of the given 
quantities y , y lf —, y^. 

It remains to find a corresponding formula for B r . In 
order to do this we return to equations (1), multiply them 

m order by 0, sin -, sin 2- -, 

J 2 m + 1 2 w + 1 

and add. This gives 



, sin z m 



rrr 



2 m + V 



r7r 



rir 



2 /// + 1 2 /// 4- 1 2 m -f- 1 

= j A„s r + a i( s„ cj + A,(.s' r , c s ) + ••• + ^,„(»s;, C„) 

All of the terms in the right member of this equation reduce 
to zero except B r S rr , which, according to Art. Ill, (17), be- 
comes equal to B r " m — Consequently we find 



(T) B r = 



2 m + 1 



TIT 



Urir 



//, sin- -4- Vo sin 2 + ••• 

. 2r7r "I 
+ y9ra sin2 W ^- n j 

(r«l, 2. 3, • ••.->. 



* It was for this reason that thp notation \ A lt . rather than A () , was chosen in 
3 for the constant term of the harmonic function. 



260 THEORY OF WAVE MOTION 

Equations (6) and (7) furnish the complete solution of 
the problem of trigonometric interpolation. The coefficients 
of the harmonic function 

y = l-A + A 1 cos x + A 2 cos 2 x -\ — 4- A m cos mx 
4- B 1 sin x + B 2 sin 2 x 4- • •• 4- # OT sin m# 
which assumes the arbitrarily assigned values 

/or £^# 2m+l equidistant values of the argument 

_ q 2 7T 4 7T 4m7T 

' 2m + 1' 2w4-l' '2m +1' 

are ^'ven % equations (6) <mc? (7). 



EXERCISE LXII 

1. Solve the problem treated in detail in Art. 110 by the method of 
Art. 112. 

2. Solve the problems of Exercise LXI by the general method of 
Art. 112. 



INDEX 

ices are to pages 



Abscissa 

Accuracy, degree o\. in cal- 
culation 

of live-place tables .... 

of measurements .... 
Addition theorems . 184,185, 
Ambiguous case of oblique tri- 
angles Ill- 
Amplitude 228, 

Analysis. Harmonic .... 
Angle, between directed lines 

cardinal 145, 

general 

horizontal 

inclined 

initial side of 

measurement of .... 

natural measure of . . . 

negative 

of elevation or depression . 

positive 

quadrant of an 

standard position of . . . 

subtended 

terminal side of .... 

vertical 

Arc. length of 

functions of 

Area of a triangle . . .84,9] 

Aristotle 

Aryabhata 155, 

Auxiliaries Sand T . . 61, 
Axis of a coordinate system 

Ball 

Base of a system of logarithms 
Bearing 



37 
59 

1-5 
212 

114 
232 
243 

204 
173 
133 

73 

73 

135 

3 

165 

134 

73 
134 
145 
136 

74 
135 

73 
165 
152 

17 
192 
198 

138 



188 Briggs 47 

i Biirgi 38 

Cajori 39 

Center of oscillation .... 241 

Chaining 74 

Characteristic, of a loga- 
rithm 47, 49 

Checks 30, 70 

Chord 155 

Cofunctions 21 

Cologarithm 54 

Compass 5, 74 

Complementary angles, func- 
tions of 20, 169 

Composition, of displacements, 
velocities, forces and vectors 126 

Coordinates 138, 139 

Cosecant, definition of . . 14, 140 

graph of 163 

line representation of . . . 154 

Cosecant Curve 163 

Cosine, definition of . . 14, 87, 140 

graph of 161 

line representation of . . . 152 

Cosine Curve 161 

Cosines, law of 85 

Cos (a ± 0) 184, 186 

Cos A ± cos B 188 

Cos 2 a 189 

Cotangent, definition of . . 14, 140 

graph of 163 

line representation of . . 153 

Cotangent Curve 163 

17 Cot (a ± P) 185, 186 

13 Crelle's Tables 38 

7 \ Crest of a wave 232 

261 



262 



INDEX 



Decimal places, number of . . 34 

Departure 75 

Descartes 130, 159 

Difference of latitude ... 75 
Difference of two angles, func- 
tions of 186 

Difference of two cosines . . 188 

of two sines 188 

Directed lines 201 

line segments 202 

Direction cosines 209 

Dirichlet 242 

Displacements 126 

Distances, measurements of . 1 

Double angles, functions of . 189 

Eichhorn slide rule . . . 66, 95 
Equations, trigonometric 222-225 

Errors, gross 68 

small ■ 70, 71 

Eudemus 7 

Euler 250 

Extraction of roots by loga- 
rithms 55 

Fermat 159 

Fink . .101 

Forces 127 

Form ratios of a triangle . . 98 

Foster 63 

Fourier 242 

Framework of a calculation . 69 

Fuller's slide rule 65 

Function, definition of . . . 13 

graph of 156-163 

Functions, of acute angles . 14 

of cardinal angles . . . 147 

- of complementary angles 20, 169 

of double angles . . . . 189 

of a general angle . 137, 140 

of half angles 190 

of obtuse angle .... 89 
of small angles or angles 
near a right angle ... 60 



of sum or difference . 184, 186 

of supplementary angles . 89 

of symmetric angles . . . 167 

Grade . . . 12 

Gradient 12 

Graphic method 8 

Graphs, of functions, some of 

whose values are given . 156 
of simple algebraic func- 
tions 158 

of the trigonometric func- 
tions 159-163 

Greek alphabet 132 

Gross errors 68 

Gunter 62 

Gunter's scale 61 

Half angle formulas .... 95 

Half angle, functions of . . 190 

Harmonic analysis 243 

Harmonic curves, .simple . . 231 

general 241 

Harmonic motion, simple . . 226 

general 239 

Heights and distances . . . 118 

Hero 92 

Hero's formula 92 

Hipparchus 192 

Horizontal angles, lines and 

planes 73 

Identities, involving a single 

angle 149 

involving several angles . 185 

Inaccuracies of measurements . 1 
Inclined lines, planes and 

angles 73 

Index of Refraction .... 1 

Index laws 41 

Indices, fractional, negative 

and zero 40 

Inskip's Tables 31 



ixnr.x 



•J«;.-; 



Interpolation, in ose of 
trigonometric 

Inverse functions 



tablet 



Latitude .... 

Law. of cosines 
of sines . . . 
of sines generalized 
of tangents . . 

Level 

Level chaining . . 

Leveling .... 

Limits of variation of the 
tions . . . 



T . . sin , tan 
Limits and 



II. 9< 



func 



Line representation of func- 
tions 151 

Lines, directed 201 

Line segments, directed . . . 202 
Logarithms, arrangement and 

use of the tables . . 51-53 
base of a system .... 43 
calculation with .... 47 
characteristics of . . . 47. 49 

common 47 

definition of 43 

extraction of roots by . . 55 

mantissa of 47. 48 

of numbers 47-50 

of trigonometric functions 57 
properties of .... 14, 45 
Logarithmic calculations in- 
volving negative numbers 56 

Logarithmic scale 61 

Longitude 78 



26 Meridian 71 

24:5 Method of mathematical indue - 

217 tion 170. 184 

JMichelson 243 

Mobius 213 

Mollweide 105 

Mollweide's equations . . . 104 

Napier 38 

Nasir Addin 84 

Natural functions 25 

Natural unit of circular meas- 
urement 165 

Navigation 75 

Negative angles 134 

line segments 202 

Newton 105 



85 

212 

212 

101 

5 

74 

147 
195 



Mannheim .... 
Mannheim's slide rule 
Mantissa .... 
Mariner's compass 
Measurement of lines 

angles .... 
Menelaus .... 



and 



Object of trigonometry ... 10 
Oblique triangles, ambiguous 

case of 111-114 

area of 84, 91, 92 

solution of 107-131 

theory of 82-106 

Ordinate 138 

Origin of arcs, primary and 

secondary 151, 152 

Origin of coordinates .... 138 
Oscillation, center of ... . 241 
Oughtred 62 

Parallax 198 

Parallelogram of forces . . . 127 
Periodicity of trigonometric 

functions 147 

Period of a simple harmonic 

motion 228 

of a wave motion .... 238 

Phase 229 

Phase constant . . . 228, 238, 239 

75 Pitiscus 192 

Plane sailing 118 

1,3 Plane surveying .... 73,118 

1 92 Plumb line 72 



. 64 
. 63 

47,48 



264 



INDEX 



Plutarch 7 

Points of the compass ... 75 
Pothenot's problem .... 123 
Principal values of inverse func- 
tions 219 

Products of sines and cosines . 187 

Profile of a wave 236 

Projection, definition of . . . 204 

theorems on 207 

of a broken line .... 208 

Proportional parts, tables of . 53 

Protractor 3 

Ptolemy 87, 192 

Pythagoras, generalized theo- 
rem of 85 

Quadrantal formulae . . 171-173 

Quadrants, the four .... 145 

Radian 165 

Radius, of circumscribed circle 96 

of escribed circle .... 97 

of inscribed circle .... 94 

Radius vector 139 

Ratios, trigonometric functions 

defined as 15 

Reflection of light 128 

Refraction of light .... 129 
Relations, between functions of 

a single angle . . .IS, 148 
between functions of com- 
plementary angles . . 20, 169 
between functions of supple- 
mentary angles .... 89 
between functions of sym- 
metric angles 167 

Rheticus 192 

Right triangles, of unfavorable 

dimensions 80 

solution by logarithms . . 67 
solution by natural func- 
tions 29 

S, the auxiliary . . . . 61, 198 

Secant curve ....... 163 



Secant, definition of ... . 14 

graph of . . 163 

line representation of . . 154 

Sector, area of 166 

Segment, area of 166 

Sense of a line segment . . . 202 
Signs of coordinates .... -13iL_ 

of trigonometric functions . 145 
Sine, definition of . . . 14, 83, 140 

graph of . 161 

line representation of . . 152 

origin of name 155 

Sine curve 161 

Sines, law of . . . 41, 96, 212 

Sin (a ± p) 184, 186 

Sin 2 a 189 

Sin A ± Sin B 100 

Skeleton form of a computation 69 

Slide rule . 62 

Slope 12 

Small angles, functions of . . 60 

Smoley's tables 31 

Snellius 125, 130, 252 

Solution of oblique triangles 107-131 
of right triangles ... 29, 67 

Spirit level 5 

Squares, table of 30 

Standard position of an angle . 136 

Stratton 243 

Subtended angle 74 

Subtraction formulae .... 186 
Sum of two angles, functions 

of 184, 185,212 

Sum of two sines or cosines . 188 
Supplementary angles, func- 
tions of 89 

Symmetric angles, functions of 167 

T, the auxiliary .... 61, 198 
Tables, explanation of . 25, 30, 57 
Tangent, definition of . . 14, 140 

graph of 163 

line representation of . . 153 
Tangent curve 163 



INDEX 



265 



Tangents, law of 101 

Tan (oiP) 185, 186 

Thacher's slide rule .... 

Thales of Miletus 

Theodolite 

Transit 

Trigonometric functions, defini- 
tions of . . . . 14, 89, 
table of logarithms of 



table of values of . . 
Trigonometric interpolation 
Trigonometry, object of . 
Trough of a wave . . . 



65 

7 
1 
4 

llo 
:>7 



•Jo 
243 

10 
232 



Variation of functions 



146 



Vector . 127 

Velocities 127 

Velocity of propagation of a 

wave '2'-)ti 

Vernier 5 

Vertical lines, planes and 

angles 72. 7:; 

Vieta 87, 101 

Von Opel 105 

Wave length 232 

Wave motion 235 

Zero, division by 142 



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